Question

PLEASE HELP

Answer 1

First graph

• y=-16x^2+22x+3(Graph attached)

#A

Use quadratic formula

$\\ \rm\rightarrowtail x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\\ \rm\rightarrowtail x=\dfrac{-22\pm\sqrt{484+192}}{-32}$

$\\ \rm\rightarrowtail x=\dfrac{-22\pm\sqrt{676}}{-32}]/tex][tex]\\ \rm\rightarrowtail x=\dfrac{-22\pm 26}{-32}$

$\\ \rm\rightarrowtail x=\dfrac{-1}{8}\:or\:\dfrac{3}{2}$

$\\ \rm\rightarrowtail x=-0.125\:or\:1.5$

#B

Vertex will be maximum

Coordinates are

• (0.688,10.563)

#C

• We have to find x intercepts (Part A)
• Then vertex(Part B)
• Then we can graph it

Answer 2

$\sf f(x) = -16x^2 + 22x + 3$

Part A

To find x-intercepts, the f(x) will be 0

$\sf \hookrightarrow -16x^2 + 22x + 3=0$

$\sf \hookrightarrow -16x^2 + 24x -2x+ 3=0$

$\sf \hookrightarrow -8x(2x-3)-1(2x- 3)=0$

$\sf \hookrightarrow (-8x-1)(2x+ 3)=0$

$\sf \hookrightarrow -8x-1=0 , \ 2x+ 3=0$

$\sf \hookrightarrow x=-0.125 , \ x=1.5$

find y-intercept:

to find y-intercept, x will be 0

$\sf f(x) = -16(0)^2 + 22(0) + 3$

$\sf f(x) = 3$

Part B

To find vertex use the formula: x = -b/2(a)

$\sf \rightarrow x = \dfrac{-22}{2(-16)}$

$\sf \rightarrow x = 0.6875$

find y coordinates:

$\sf f(x) = -16(0.6875)^2 + 22(0.6875) + 3$

$\sf f(x) =10.5625$

coordinates of vertex: (0.6875, 10.5625)

• $\bold{\mathrm{If}\:a < 0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}}$

• $\bold{\mathrm{If}\:a > 0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}}$  

As its positive and greater than 0, the vertex will be Maximum.

part C

To draw the graph, simplify point the vertex on the graph. Then point both the x-intercepts. Also the y-intercept and draw the curve with free hand.

Graph attached: