Solved Examples on Quadratic Functions

$ Standard\;\;Form:\;\; y = -2x^2 + 8x + 11 \\[3ex] Compare:\;\; y = ax^2 + bx + c \\[3ex] a = -2,\;\; b = 8,\;\; c = 11 \\[3ex] (a.) \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] -\dfrac{b}{2a} = -\dfrac{8}{2(-2)} \\[5ex] -\dfrac{b}{2a} = \dfrac{-8}{-4} = 2 \\[5ex] f\left(-\dfrac{b}{2a}\right) = f(2) \\[5ex] But\:\: f(x) = -2x^2 + 8x + 11 \\[3ex] f(2) = -2(2)^2 + 8(2) + 11 \\[3ex] f(2) = -2(4) + 16 + 11 \\[3ex] f(2) = -8 + 16 + 11 \\[3ex] f(2) = 19 \\[3ex] Vertex = (2, 19) \\[3ex] (b.) \\[3ex] \underline{1st\;\;Method:\;\;Vertex\;\;Formula} \\[3ex] Vertex\:\: Form:\:\: y = a(x - h)^2 + k \\[3ex] Vertex = (2, 19) \\[3ex] Vertex = (h, k) \\[3ex] h = 2 \\[3ex] k = 19 \\[3ex] Vertex\:\: Form:\:\: y = -2(x - 2)^2 + 19 \\[3ex] \underline{2nd\;\;Method:\;\;Completing\;\;the\;\;Square\;\;Method} \\[3ex] Standard\;\;Form:\;\; y = -2x^2 + 8x + 11 \\[3ex] f(x) = -2\left(x^2 - \dfrac{8x}{2} - \dfrac{11}{2}\right) \\[5ex] f(x) = -2\left(x^2 - 4x - \dfrac{11}{2}\right) \\[5ex] Coefficient\;\;of\;\;x = -4 \\[3ex] Half\;\;of\;\;it = \dfrac{1}{2} * (-4) = -2 \\[5ex] Square\;\;it = (-2)^2 \\[3ex] f(x) = -2\left[x^2 - 4x + (-2)^2 - \dfrac{11}{2} - (-2)^2\right] \\[5ex] f(x) = -2\left[(x - 2)^2 - \dfrac{11}{2} - 4\right] \\[5ex] f(x) = -2\left[(x - 2)^2 - \dfrac{11}{2} - \dfrac{8}{2}\right] \\[5ex] f(x) = -2\left[(x - 2)^2 - \dfrac{19}{2}\right] \\[5ex] f(x) = -2(x - 2)^2 + 19 \\[5ex] $ To find the $y-intercept$, set $x = 0$ and solve for $y$

$ (c.) \\[3ex] y = -2x^2 + 8x + 11 \\[3ex] x = 0 \\[3ex] y = -2(0)^2 + 8(0) + 11 \\[3ex] y = -2(0) + 0 + 11 \\[3ex] y = 0 + 0 + 11 = 11 \\[3ex] y-intercept = (0, 11) \\[3ex] $ To find the $x-intercept$, set $y = 0$ and solve for $x$

$ (d.) \\[3ex] y = -2x^2 + 8x + 11 \\[3ex] y = 0 \\[3ex] 0 = -2x^2 + 8x + 11 \\[3ex] -2x^2 + 8x + 11 = 0 \\[3ex] a = -2, b = 8, c = 11 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-8 \pm \sqrt{(-8)^2 - 4(-2)(11)}}{2(-2)} \\[5ex] x = \dfrac{-8 \pm \sqrt{64 + 88}}{-4} \\[5ex] x = \dfrac{-8 \pm \sqrt{152}}{-4} \\[5ex] x = \dfrac{-8 \pm \sqrt{4 * 38}}{-4} \\[5ex] x = \dfrac{-8 \pm \sqrt{4} * \sqrt{38}}{-4} \\[5ex] x = \dfrac{-8 \pm 2 * \sqrt{38}}{-4} \\[5ex] x = \dfrac{-2\left(4 \mp \sqrt{38}\right)}{-4} \\[5ex] x = \dfrac{4 \mp \sqrt{38}}{2} \\[5ex] x = \dfrac{4 - \sqrt{38}}{2} \:\:\:OR\:\:\: x = \dfrac{4 + \sqrt{38}}{2} \\[5ex] x-intercepts = \left(\dfrac{4 - \sqrt{38}}{2}, 0\right) \:\:and\:\: \left(\dfrac{4 + \sqrt{38}}{2}, 0\right) $

$ y = 30(x + 17)^2 - 42 \\[3ex] Compare to \\[3ex] Vertex\:\:Form\:\:of\:\:Quadratic\:\:Function \\[3ex] y = a(x - h)^2 + k \\[3ex] -h = 17 \\[3ex] \rightarrow h = -17 \\[3ex] k = -42 \\[3ex] Vertex = (h, k) \\[3ex] \therefore Vertex = (-17, -42) $

$ (a.) \\[3ex] f(x) = x^2 + 4x \\[3ex] Compare \\[3ex] f(x) = ax^2 + bx + c \\[3ex] a = 1 \\[3ex] 1 \gt 0 \\[3ex] Graph\;\;opens\;\;up \\[3ex] (b.) \\[3ex] a = 1 \\[3ex] b = 4 \\[3ex] x-coordinate\;\;of\;\;vertex = -\dfrac{b}{2a} = -\dfrac{4}{2(1)} = \dfrac{-4}{2} = -2 \\[5ex] y-coordinate\;\;of\;\;vertex = f(-2) = (-2)^2 + 4(-2) = 4 - 8 = -4 \\[3ex] vertex = \left(-\dfrac{b}{2a}, f(-\dfrac{b}{2a})\right) = (-2, -4) \\[5ex] (c.) \\[3ex] axis\;\;of\;\;symmetry:\;\; x = h \\[3ex] x = -2 \\[3ex] (d.) \\[3ex] \underline{x-intercept} \\[3ex] set\;\;y = 0\;\;and\;\;solve\;\;for\;\;x \\[3ex] f(x) = x^2 + 4x \\[3ex] x^2 + 4x = f(x) \\[3ex] x^2 + 4x = y \\[3ex] x^2 + 4x = 0 \\[3ex] x(x + 4) = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x + 4 = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x = -4 \\[3ex] x-intercepts = (0, 0) \;\;\;and\;\;\; (-4, 0) \\[3ex] \underline{y-intercept} \\[3ex] set\;\;x = 0\;\;and\;\;solve\;\;for\;\;y \\[3ex] f(0) = 0^2 + 4(0) \\[3ex] f(0) = 0 + 0 = 0 \\[3ex] y-intercept = (0, 0) \\[3ex] $ (e.) To graph this function in MML (MyMath Lab), we need two points.
The vertex is one point.
One of the x-intercepts is another point.
The graph of the quadratic function is:



(f.) From this graph:
The domain includes all real numbers of x
The lowest y-value is −4. It is included. It is the minimum value of y
There is no maximum value for y
Therefore, the range includes all real numbers of y from −4 (included) to ∞

$ D = (-\infty, \infty) \\[3ex] R = [-4, \infty) \\[3ex] (g.) \\[3ex] f(x) \uparrow \;\;for\;\; x \in (-2, \infty) \\[3ex] f(x) \downarrow \;\;for\;\; x \in (-\infty, -2) $

$ (a.) \\[3ex] f(x) = -x^2 - 4x \\[3ex] Compare \\[3ex] f(x) = ax^2 + bx + c \\[3ex] a = -1 \\[3ex] -1 \lt 0 \\[3ex] Graph\;\;opens\;\;down \\[3ex] (b.) \\[3ex] a = -1 \\[3ex] b = -4 \\[3ex] x-coordinate\;\;of\;\;vertex = -\dfrac{b}{2a} = -\dfrac{-4}{2(-1)} = \dfrac{4}{-2} = -2 \\[5ex] y-coordinate\;\;of\;\;vertex = f(-2) = -(-2)^2 - 4(-2) = -4 + 8 = 4 \\[3ex] vertex = \left(-\dfrac{b}{2a}, f(-\dfrac{b}{2a})\right) = (-2, 4) \\[5ex] (c.) \\[3ex] axis\;\;of\;\;symmetry:\;\; x = h \\[3ex] x = -2 \\[3ex] (d.) \\[3ex] \underline{x-intercept} \\[3ex] set\;\;y = 0\;\;and\;\;solve\;\;for\;\;x \\[3ex] f(x) = -x^2 - 4x \\[3ex] -x^2 - 4x = f(x) \\[3ex] -x^2 - 4x = y \\[3ex] -x^2 - 4x = 0 \\[3ex] -x(x + 4) = 0 \\[3ex] -x = 0 \;\;\;OR\;\;\; x + 4 = 0 \\[3ex] x = 0 \;\;\;OR\;\;\; x = -4 \\[3ex] x-intercepts = (0, 0) \;\;\;and\;\;\; (-4, 0) \\[3ex] \underline{y-intercept} \\[3ex] set\;\;x = 0\;\;and\;\;solve\;\;for\;\;y \\[3ex] f(0) = -0^2 - 4(0) \\[3ex] f(0) = -0 - 0 = 0 \\[3ex] y-intercept = (0, 0) \\[3ex] $ (e.) To graph this function in MML (MyMath Lab), we need two points.
The vertex is one point.
One of the x-intercepts is another point.
The graph of the quadratic function is:



(f.) From this graph:
The domain includes all real numbers of x
The highest y-value is 4. It is included. It is the maximum value of y
There is no minimum value for y
Therefore, the range includes all real numbers of y from −∞ to 4 (included)

$ D = (-\infty, \infty) \\[3ex] R = (-\infty, 4] \\[3ex] (g.) \\[3ex] f(x) \uparrow \;\;for\;\; x \in (-\infty, -2) \\[3ex] f(x) \downarrow \;\;for\;\; x \in (-2, \infty) $

$ \underline{1st\;\;Approach:\;\;Find\;\;the\;\;quadratic\;\;function} \\[3ex] Point\;1:\;\;(1, 17) \implies n = 1,\;\;d = 17 \\[3ex] 17 = a(1)^2 + b(1) + c \\[3ex] 17 = a(1) + b + c \\[3ex] 17 = a + b + c \\[3ex] a + b + c = 17...eqn.(1) \\[3ex] Point\;2:\;\;(2, 10) \implies n = 2,\;\;d = 10 \\[3ex] 10 = a(2)^2 + b(2) + c \\[3ex] 10 = a(4) + 2b + c \\[3ex] 10 = 4a + 2b + c \\[3ex] 4a + 2b + c = 10...eqn.(2) \\[3ex] Point\;3:\;\;(3, 5) \implies n = 3,\;\;d = 5 \\[3ex] 5 = a(3)^2 + b(3) + c \\[3ex] 5 = a(9) + 3b + c \\[3ex] 5 = 9a + 3b + c \\[3ex] 9a + 3b + c = 5...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \rightarrow \\[3ex] 4a + 2b + c - (a + b + c) = 10 - 17 \\[3ex] 4a + 2b + c - a - b - c = -7 \\[3ex] 3a + b = -7...eqn.(4) \\[3ex] eqn.(3) - eqn.(2) \rightarrow \\[3ex] 9a + 3b + c - (4a + 2b + c) = 5 - 10 \\[3ex] 9a + 3b + c - 4a - 2b - c = -5 \\[3ex] 5a + b = -5...eqn.(5) \\[3ex] eqn.(5) - eqn.(4) \rightarrow \\[3ex] 5a + b - (3a + b) = -5 - (-7) \\[3ex] 5a + b - 3a - b = -5 + 7 \\[3ex] 2a = 2 \\[3ex] a = \dfrac{2}{2} \\[5ex] a = 1 \\[3ex] From\;\;eqn.(4) \\[3ex] 3a + b = -7...eqn.(4) \\[3ex] b = -7 - 3a \\[3ex] b = -7 - 3(1) \\[3ex] b = -7 - 3 \\[3ex] b = -10 \\[3ex] From\;\;eqn.(1) \\[3ex] a + b + c = 17...eqn.(1) \\[3ex] c = 17 - a - b \\[3ex] c = 17 - 1 - (-10) \\[3ex] c = 17 - 1 + 10 \\[3ex] c = 26 \\[3ex] d = an^2 + bn + c \\[3ex] d = 1n^2 + (-10)n + 26 \\[5ex] Quadratic\;\;Function:\;\; d = n^2 - 10n + 26 \\[3ex] $ ##########
$ If\;\;you\;\;have\;\;time \\[3ex] Check\;\;for\;\;Point\;\;4\;\;to\;\;confirm \\[3ex] Point\;4:\;\;(4, 2) \implies n = 4,\;\;d = 2 \\[3ex] d = n^2 - 10n + 26 \\[3ex] d = 4^2 - 10(4) + 26 \\[3ex] d = 16 - 40 + 26 \\[3ex] d = 2...confirmed $ ##########

$ (5.1) \\[3ex] Point\;5:\;\;(5, r) \implies n = 5,\;\;d = r \\[3ex] r = 5^2 - 10(5) + 26 \\[3ex] r = 25 - 50 + 26 \\[3ex] r = 1 \\[3ex] Point\;6:\;\;(6, s) \implies n = 6,\;\;d = s \\[3ex] s = 6^2 - 10(6) + 26 \\[3ex] s = 36 - 60 + 26 \\[3ex] s = 2 \\[3ex] \underline{2nd\;\;Approach:\;\;Quadratic\;\;Sequence\;\;Approach} \\[3ex] Quadratic\;\;Sequence = 17, 10, 5, 2, r, s \\[3ex] Quadratic\;\;Sequence = 1st,\;\; 2nd,\;\; 3rd,\;\; 4th,\;\; 5th,\;\; 6th \\[3ex] \underline{1st\;\;Difference} \\[3ex] 10 - 17 = -7 \\[3ex] 5 - 10 = -5 \\[3ex] 2 - 5 = -3 \\[3ex] r - 2 = r - 2 \\[3ex] s - r = s - r \\[3ex] \underline{2nd\;\;Difference} \\[3ex] -5 - (-7) = -5 + 7 = 2 \\[3ex] -3 - (-5) = -3 + 5 = 2 \\[3ex] r - 2 - (-3) = r - 2 + 3 = r + 1 \\[3ex] s - r - (r - 2) = s - r - r + 2 = s - 2r + 2 \\[3ex] 2nd\;\;Differences\;\;must\;\;be\;\;the\;\;same \\[3ex] \implies \\[3ex] 2 = 2 \\[3ex] r + 1 = 2 \\[3ex] r = 2 - 1 \\[3ex] r = 1 \\[3ex] s - 2r + 2 = 2 \\[3ex] s - 2(1) + 2 = 2 \\[3ex] s - 2 + 2 = 2 \\[3ex] s = 2 \\[3ex] \underline{nth\;\;term\;\;of\;\;a\;\;Quadratic\;\;Sequence} \\[3ex] QS_n = \dfrac{1st + 3rd - 2(2nd)}{2} * n^2 + \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} * n + 3(1st) - 3(2nd) + 3rd \\[5ex] QS_n = \dfrac{17 + 5 - 2(10)}{2} * n^2 + \dfrac{8(10) - 5(17) - 3(5)}{2} * n + 3(17) - 3(10) + 5 \\[5ex] QS_n = \dfrac{17 + 5 - 20}{2} * n^2 + \dfrac{80 - 85 - 15}{2} * n + 51 - 30 + 5 \\[5ex] QS_n = \dfrac{2}{2} * n^2 + \dfrac{-20}{2} * n + 26 \\[5ex] QS_n = n^2 - 10n + 26 \\[3ex] \implies d = n^2 - 10n + 26 \\[3ex] (5.2) \\[3ex] a = 1,\;\; b = -10,\;\; c = 26 \\[3ex] (5.3) \\[3ex] When\;\;n = 8,\;\;what\;\;is\;\;d? \\[3ex] d = n^2 - 10n + 26 \\[3ex] d = 8^2 - 10(8) + 26 \\[3ex] d = 64 - 80 + 26 \\[3ex] d = 10\;metres \\[3ex] (5.4) \\[3ex] \underline{By\;\;Inspection} \\[3ex] Vertex = Minimum\;\;Point = (5, r) = (5, 1) \\[3ex] This\;\;is\;\;because\;\;the\;\;next\;\;distance:\;\;s = 2...starts\;\;increasing \\[3ex] \underline{By\;\;Standard\;\;Form\;\;Computation} \\[3ex] d = n^2 - 10n + 26 \\[3ex] Compare:\;\; y = ax^2 + bx + c \\[3ex] a = 1,\;\; b = -10,\;\; c = 26 \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] -\dfrac{b}{2a} = -\dfrac{-10}{2(1)} = \dfrac{10}{2} = 5 \\[5ex] f\left(-\dfrac{b}{2a}\right) = f(5) = d(5) = 1 \\[3ex] The\;\;Quadratic\;\;Function\;\;is: \\[3ex] (time, distance) = ((1, 17),\;\;(2, 10)\;\;(3, 5)\;\;(4, 2)\;\;(5, 1)\;\;(6, 2)\;\;(7, 5)\;\;(8, 10)\;\;(9, 17) \\[3ex] $ Based on the Table of Values or the value pairs, we see that:
As the time approaches five seconds $(n \lt 5)$, the distance from the fixed point, $P$ decreases. This indicates that the athlete is moving towards $P$
After five seconds $(n \gt 5)$, the distance from the fixed point, $P$ starts increasing. This indicates that the athlete is moving away from $P$

Another approach is to use Calculus

$ d = n^2 - 10n + 26 \\[3ex] \dfrac{dd}{dn} = 2n - 10 \\[3ex] When\;\;n \lt 5: \\[3ex] 2n - 10 \lt 0 \\[3ex] \dfrac{dd}{dn} \lt 0...decreasing \\[3ex] When\;\;n \gt 5: \\[3ex] 2n - 10 \gt 0 \\[3ex] \dfrac{dd}{dn} \gt 0...increasing $

$ y = ax^2 + bx + 2 \\[3ex] 1st\;\;Point:\;\; (-1, 13) \implies x = -1, y = 13 \\[3ex] 13 = a(-1)^2 + b(-1) + 2 \\[3ex] 13 = a(1) - b + 2 \\[3ex] 13 = a - b + 2 \\[3ex] a - b + 2 = 13 \\[3ex] a - b = 13 - 2 \\[3ex] a - b = 11...eqn.(1) \\[3ex] 2nd\;\;Point:\;\; (4, 18) \implies x = 4, y = 18 \\[3ex] 18 = a(4)^2 + b(4) + 2 \\[3ex] 18 = 16a + 4b + 2 \\[3ex] 16a + 4b + 2 = 18 \\[3ex] 16a + 4b = 18 - 2 \\[3ex] 16a + 4b = 16 \\[3ex] Simplify:\;\;Divide\;\;all\;\;sides\;\;by\;\;4 \\[3ex] 4a + b = 4...eqn.(2) \\[3ex] To\;\;find\;\;a \\[3ex] eqn.(2) + eqn.(1) \rightarrow \\[3ex] (4a + b) + (a - b) = 4 + 11 \\[3ex] 4a + b + a - b = 15 \\[3ex] 5a = 15 \\[3ex] a = \dfrac{15}{5} \\[5ex] a = 3 \\[3ex] From\;\;eqn.(1) \\[3ex] a - b = 11 \\[3ex] a - 11 = b \\[3ex] 3 - 11 = b \\[3ex] -8 = b \\[3ex] b = -8 \\[3ex] \underline{Check}...if\;\;you\;\;have\;\;time \\[3ex] y = ax^2 + bx + 2 \\[3ex] y = 3x^2 - 8x + 2 \\[3ex] When\;\; x = -1 \\[3ex] y = 3(-1)^2 - 8(-1) + 2 \\[3ex] y = 3(1) + 8 + 2 \\[3ex] y = 3 + 8 + 2 \\[3ex] y = 13 \\[3ex] works\;\;for\;\;(-1, 13)...1st\;\;point \\[3ex] When\;\; x = 4 \\[3ex] y = 3(4)^2 - 8(4) + 2 \\[3ex] y = 3(16) - 32 + 2 \\[3ex] y = 48 - 32 + 2 \\[3ex] y = 18 \\[3ex] works\;\;for\;\;(4, 18)...2nd\;\;point $

$ (i) \\[3ex] From\;\;the\;\;graph \\[3ex] x = 2 \;\;\;OR\;\;\; x = 4 \\[3ex] \underline{Check\;\;Algebraically\;\;if\;\;you\;\;have\;\;time} \\[3ex] x^2 - 6x + 8 = 0 \\[3ex] \underline{LHS} \\[3ex] x^2 - 6x + 8 \\[3ex] x = 2 \\[3ex] 2^2 - 6(2) + 8 \\[3ex] 4 - 12 + 8 \\[3ex] = 0 = RHS \\[3ex] \underline{LHS} \\[3ex] x^2 - 6x + 8 \\[3ex] x = 4 \\[3ex] 4^2 - 6(4) + 8 \\[3ex] 16 - 24 + 8 \\[3ex] = 0 = RHS \\[3ex] (ii) \\[3ex] Minimum\;\;point = Vertex = (3, -1) \\[3ex] \underline{Check\;\;Algebraically\;\;if\;\;you\;\;have\;\;time} \\[3ex] x^2 - 6x + 8 = 0 \\[3ex] Compare:\;\; ax^2 + bx + c = 0 \\[3ex] a = 1, \;\;\; b = -6, \;\;\; c = 8 \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] -\dfrac{b}{2a} = -\dfrac{-6}{2(1)} = \dfrac{6}{2} = 3 \\[5ex] f\left(-\dfrac{b}{2a}\right) = f(3) \\[3ex] f(x) = x^2 - 6x + 8 \\[3ex] f(3) = 3^2 - 6(3) + 8 \\[3ex] f(3) = 9 - 18 + 8 \\[3ex] f(3) = -1 \\[3ex] \implies Vertex = (3, -1) \\[3ex] (iii) \\[3ex] Two\;\;methods\;\;to\;\;solve\;\;this\;\;question \\[3ex] Use\;\;whichever\;\;method\;\;you\;\;prefer \\[3ex] \underline{1st\;\;Method:\;\;Vertex\;\;Formula} \\[3ex] Vertex = (3, -1) \\[3ex] Vertex = (h, k) \\[3ex] \implies (h, k) = (3, -1) \\[3ex] h = 3, \;\;\; k = -1 \\[3ex] Vertex\;\;Form = a(x - h)^2 + k \\[3ex] Vertex\;\;Form = 1(x - 3)^2 + (-1) \\[3ex] Vertex\;\;Form = (x - 3)^2 - 1 \\[3ex] \underline{2nd\;\;Method:\;\;Completing\;\;the\;\;Square\;\;Method} \\[3ex] x^2 - 6x + 8 \\[3ex] Coefficient\;\;of\;\;x = -6 \\[3ex] Half\;\;of\;\;it = \dfrac{1}{2} * -6 = -3 \\[5ex] Square\;\;it = (-3)^2 \\[3ex] \implies \\[3ex] x^2 - 6x + (-3)^2 + 8 - (-3)^2 \\[3ex] (x - 3)^2 + 8 - 9 \\[3ex] (x - 3)^2 - 1 \\[3ex] (iv) \\[3ex] $
$ (v) \\[3ex] x = 2 \;\;\;OR\;\;\; x = 5 \\[3ex] $

Check if you have time

$LHS$	$RHS$
$x = 2$ $x^2 - 6x + 8$ $2^2 - 6(2) + 8$ $4 - 12 + 8$ $0$	$x = 2$ $x - 2$ $2 - 2$ $0$
$x = 5$ $x^2 - 6x + 8$ $5^2 - 6(5) + 8$ $25 - 30 + 8$ $3$	$x = 5$ $x - 2$ $5 - 2$ $3$

$ (a.) \\[3ex] g(x) = -2(x - 3)^2 + 1 \\[3ex] Compare \\[3ex] g(x) = a(x - h)^2 + k \\[3ex] h = 3 \\[3ex] k = 1 \\[3ex] vertex = (h, k) = (3, 1) \\[3ex] (b.) \\[3ex] axis\;\;of\;\;symmetry:\;\; x = h \\[3ex] x = 3 \\[3ex] (c.) \\[3ex] a = -2 \\[3ex] -2 \lt 0 \\[3ex] Graph\;\;is\;\;concave\;\;down \\[3ex] $ To graph this function in MML (MyMath Lab), we need two points.
The vertex is one point.
Let us find the y-intercept as the second point

$ g(0) = -2(0 - 3)^2 + 1 \\[3ex] g(0) = -2(-3)^2 + 1 \\[3ex] g(0) = -2(9) + 1 \\[3ex] g(0) = -18 + 1 \\[3ex] g(0) = -17 \\[3ex] y-intercept = (0, -17) \\[3ex] $ (d.) The graph of the quadratic function is:

(a.)

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$x^2$	$16$	$9$	$4$	$1$	$0$	$1$	$4$	$9$	$16$
$\dfrac{1}{2}$	$0.5$	$0.5$	$0.5$	$0.5$	$0.5$	$0.5$	$0.5$	$0.5$	$0.5$
$y = x^2 + \dfrac{1}{2}$	$16.5$	$9.5$	$4.5$	$1.5$	$0.5$	$1.5$	$4.5$	$9.5$	$16.5$

(b.) Draw the graph according to the scale specified in the question.

$ (c.) \\[3ex] (i) \\[3ex] y = x^2 + \dfrac{1}{2} \\[5ex] x^2 = 5 \\[3ex] y = 5 + \dfrac{1}{2} \\[5ex] y = 5 + 0.5 \\[3ex] y = 5.5 \\[3ex] $ Draw the graph of $y = 5.5$...this is the graph of a constant. It is a straight line graph
Find the intersection of the straight line graph and the parabola (graph of the quadratic function)
The point of intersection is the solution of $x^2 = 5$

To find the range of values of $x$ for which $y$ decreases as $x$ decreases, let us first find the vertex: the minimum point on the graph (in this case).
This is because it is the point at which the graph cuts in half, and the $y-values$ begin to repeat

$ (ii) \\[3ex] y = x^2 + \dfrac{1}{5} \\[5ex] Compare\;\;to\;\; y = ax^2 + bx + c \\[3ex] a = 1 \\[3ex] b = 0 \\[3ex] c = \dfrac{1}{5} \\[5ex] x-component\;\;of\;\;the\;\;vertex \\[3ex] = -\dfrac{b}{2a} \\[5ex] = -\dfrac{0}{2(1)} \\[5ex] = 0 \\[3ex] $ Based on the graph, the $y-values$ begin to decrease from $x = -4$ up to the minimum point (vertex).
Therefore, the range of values of $x$ for which $y$ decreases as $x$ decreases is $[-4, 0]$

$ (a.) \\[3ex] g(x) = 2(x - 6)^2 + 5 \\[3ex] Compare \\[3ex] g(x) = a(x - h)^2 + k \\[3ex] h = 6 \\[3ex] k = 5 \\[3ex] vertex = (h, k) = (6, 5) \\[3ex] (b.) \\[3ex] axis\;\;of\;\;symmetry:\;\; x = h \\[3ex] x = 6 \\[3ex] (c.) \\[3ex] a = 2 \\[3ex] 2 \gt 0 \\[3ex] Graph\;\;is\;\;concave\;\;up \\[3ex] $ To graph this function in MML (MyMath Lab), we need two points.
The vertex is one point.
Let us find the second point

$ let\;\;x = 7 \\[3ex] g(7) = 2(7 - 6)^2 + 5 \\[3ex] g(7) = 2(1)^2 + 5 \\[3ex] g(7) = 2(1) + 5 \\[3ex] g(7) = 2 + 5 \\[3ex] g(7) = 7 \\[3ex] 2nd\;\;point = (7, 7) \\[3ex] $ (d.) The graph of the quadratic function is:

$ (a.) \\[3ex] f(x) = -\dfrac{1}{3}\left(x - \dfrac{1}{5}\right)^2 - \dfrac{7}{6} \\[5ex] Compare \\[3ex] f(x) = a(x - h)^2 + k \\[3ex] h = \dfrac{1}{5} \\[5ex] k = -\dfrac{7}{6} \\[5ex] vertex = (h, k) = \left(\dfrac{1}{5}, -\dfrac{7}{6}\right) \\[5ex] (b.) \\[3ex] axis\;\;of\;\;symmetry:\;\; x = h \\[3ex] x = \dfrac{1}{5} \\[5ex] (c.) \\[3ex] a = -\dfrac{1}{3} \\[5ex] -\dfrac{1}{3} \lt 0 \\[5ex] Graph\;\;is\;\;concave\;\;down \\[3ex] $ To choose the graph of this function in MML (MyMath Lab), we need two points.
We shall use the process of elimination.
The graph opens downward.
So, options A. and D. are eliminated.
Also, looking at the vertex, the value of y is negative.
Option C. is eliminated.
That leaves us with Option B. as the answer.
(d.) The graph of the quadratic function is:

$ Standard\;\;Form:\;\; y = 2x^2 - 5x + 3 \\[3ex] Compare:\;\; y = ax^2 + bx + c \\[3ex] a = 2,\;\; b = -5,\;\; c = 3 \\[3ex] \underline{1st\;\;Method:\;\;Vertex\;\;Formula} \\[3ex] Vertex = \left[-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right] \\[5ex] Vertex = (h, k) \\[3ex] h = -\dfrac{b}{2a} = -\dfrac{-5}{2(2)} = \dfrac{5}{4} \\[5ex] k = f\left(-\dfrac{b}{2a}\right) = f\left(\dfrac{5}{4}\right) \\[5ex] f(x) = 2x^2 - 5x + 3 \\[3ex] k = f\left(\dfrac{5}{4}\right) = 2\left(\dfrac{5}{4}\right)^2 - 5\left(\dfrac{5}{4}\right) + 3 \\[5ex] k = 2\left(\dfrac{5}{4}\right)\left(\dfrac{5}{4}\right) - \dfrac{25}{4} + \dfrac{3}{1} \\[5ex] k = \dfrac{25}{8} - \dfrac{25}{4} + \dfrac{3}{1} \\[5ex] k = \dfrac{25 - 50 + 24}{8} \\[5ex] k = -\dfrac{1}{8} \\[5ex] Vertex = (h, k) = \left(\dfrac{5}{4}, -\dfrac{1}{8}\right) \\[5ex] Vertex\:\: Form:\:\: y = a(x - h)^2 + k \\[3ex] Vertex\:\: Form:\:\: y = 2\left(x - \dfrac{5}{4}\right)^2 - \dfrac{1}{8} \\[5ex] \underline{2nd\;\;Method:\;\;Completing\;\;the\;\;Square\;\;Method} \\[3ex] Standard\;\;Form:\;\; y = 2x^2 - 5x + 3 \\[3ex] y = 2\left(x^2 - \dfrac{5}{2}x + \dfrac{3}{2}\right) \\[5ex] Coefficient\;\;of\;\;x = -\dfrac{5}{2} \\[5ex] Half\;\;of\;\;it = \dfrac{1}{2} * -\dfrac{5}{2} = -\dfrac{5}{4} \\[5ex] Square\;\;it = \left(-\dfrac{5}{4}\right)^2 \\[5ex] y = 2\left[x^2 - \dfrac{5}{2}x + \left(-\dfrac{5}{4}\right)^2 + \dfrac{3}{2} - \left(-\dfrac{5}{4}\right)^2\right] \\[5ex] y = 2\left[\left(x - \dfrac{5}{4}\right)^2 + \dfrac{3}{2} - \dfrac{25}{16}\right] \\[5ex] y = 2\left[\left(x - \dfrac{5}{4}\right)^2 + \dfrac{24}{16} - \dfrac{25}{16}\right] \\[5ex] y = 2\left[\left(x - \dfrac{5}{4}\right)^2 + \dfrac{24 - 25}{16}\right] \\[5ex] y = 2\left[\left(x - \dfrac{5}{4}\right)^2 - \dfrac{1}{16}\right] \\[5ex] y = 2\left(x - \dfrac{5}{4}\right)^2 - 2\left(\dfrac{1}{16}\right) \\[5ex] y = 2\left(x - \dfrac{5}{4}\right)^2 - \dfrac{1}{8} \\[5ex] Compare\;\;to \\[3ex] Question\;\;Form:\;\; y = p(x + q)^2 + r \\[3ex] \implies \\[3ex] p = 2 \\[3ex] q = -\dfrac{5}{4} \\[5ex] r = -\dfrac{1}{8} \\[5ex] q + r \\[3ex] = -\dfrac{5}{4} + -\dfrac{1}{8} \\[5ex] = -\dfrac{10}{8} - \dfrac{1}{8} \\[5ex] = \dfrac{-10 - 1}{8} \\[5ex] = -\dfrac{11}{8} \\[5ex] = -1\dfrac{3}{8} $

$ \underline{Direct\;\;Solution} \\[3ex] ...Recommended\;\;in\;\;this\;\;case \\[3ex] x^2 + ax + b = (x + 2)^2 - 3 \\[3ex] x^2 + ax + b = (x + 2)(x + 2) - 3 \\[3ex] x^2 + ax + b = x^2 + 2x + 2x + 4 - 3 \\[3ex] x^2 + ax + b = x^2 + 4x + 1 \\[3ex] Compare\;\;LHS\;\;and\;\;RHS \\[3ex] x^2 = x^2 \\[3ex] ax = 4x \implies a = 4 \\[3ex] b = 1 $