Question

how to labeling the y-intercept, vertex, and the axis of symmetry?

Answer 1

Answer:

• y-intercept = (0, 0)

,

• Vertex = (1/2, 5/4)

,

• Axis of symmetry = 1/2

Explanation:

Given the function:

$f(x)=-5x^2+5x$

The vertex is:

$\begin{gathered} v_x=-\frac{b}{2a} \\ \\ \text{Where a = -5, b = 5} \\ \\ v_x=-\frac{5}{2(-5)}=\frac{1}{2} \end{gathered}$

Substitute 1/2 into the given function to find the y-coordinate of the vertex

$\begin{gathered} v_y=-5(\frac{1}{2})^2+5(\frac{1}{2}) \\ \\ =\frac{5}{4} \end{gathered}$

Therefore, the vertex is (1/2, 5/4)

Axis of symmetry:

This is -b/2a

$\frac{1}{2}$

The y-intercept is (0, 0)