First graph
#A
Use quadratic formula
[tex]\\ \rm\rightarrowtail x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{-22\pm\sqrt{484+192}}{-32}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{-22\pm\sqrt{676}}{-32}]/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{-22\pm 26}{-32}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{-1}{8}\:or\:\dfrac{3}{2}[/tex]
[tex]\\ \rm\rightarrowtail x=-0.125\:or\:1.5[/tex]
#B
Vertex will be maximum
Coordinates are
#C
[tex]\sf f(x) = -16x^2 + 22x + 3[/tex]
Part A
To find x-intercepts, the f(x) will be 0
[tex]\sf \hookrightarrow -16x^2 + 22x + 3=0[/tex]
[tex]\sf \hookrightarrow -16x^2 + 24x -2x+ 3=0[/tex]
[tex]\sf \hookrightarrow -8x(2x-3)-1(2x- 3)=0[/tex]
[tex]\sf \hookrightarrow (-8x-1)(2x+ 3)=0[/tex]
[tex]\sf \hookrightarrow -8x-1=0 , \ 2x+ 3=0[/tex]
[tex]\sf \hookrightarrow x=-0.125 , \ x=1.5[/tex]
find y-intercept:
to find y-intercept, x will be 0
[tex]\sf f(x) = -16(0)^2 + 22(0) + 3[/tex]
[tex]\sf f(x) = 3[/tex]
Part B
To find vertex use the formula: x = -b/2(a)
[tex]\sf \rightarrow x = \dfrac{-22}{2(-16)}[/tex]
[tex]\sf \rightarrow x = 0.6875[/tex]
find y coordinates:
[tex]\sf f(x) = -16(0.6875)^2 + 22(0.6875) + 3[/tex]
[tex]\sf f(x) =10.5625[/tex]
coordinates of vertex: (0.6875, 10.5625)
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:
