Question

complete the parts below.
(a) graph the parabola $y=-16x^{2}+32x$. to do so, plot five points on the parabola: the vertex, two points to the left of the right of the vertex. then click on the graph-a-function button.

Explanation:

Step1: Find vertex x-coordinate

For parabola $y=ax^2+bx+c$, vertex $x = -\frac{b}{2a}$. Here $a=-16, b=32$:
$x = -\frac{32}{2\times(-16)} = 1$

Step2: Find vertex y-coordinate

Substitute $x=1$ into the equation:
$y = -16(1)^2 + 32(1) = -16 + 32 = 16$
Vertex: $(1, 16)$

Step3: Pick left x-values (x<1)

Choose $x=0$:
$y = -16(0)^2 + 32(0) = 0$ → Point: $(0, 0)$
Choose $x=-1$:
$y = -16(-1)^2 + 32(-1) = -16 -32 = -48$ → Point: $(-1, -48)$

Step4: Pick right x-values (x>1)

Choose $x=2$:
$y = -16(2)^2 + 32(2) = -64 + 64 = 0$ → Point: $(2, 0)$
Choose $x=3$:
$y = -16(3)^2 + 32(3) = -144 + 96 = -48$ → Point: $(3, -48)$

Answer:

Five points to plot:

1. Vertex: $(1, 16)$
2. Left of vertex: $(0, 0)$, $(-1, -48)$
3. Right of vertex: $(2, 0)$, $(3, -48)$

Plot these points and connect them to graph the parabola.