Question

identify the axis of symmetry and the vertex of the function. y=-5x^2+x+4

Answer 1

The general equation of parabole with vertex (h,k) is,

$y=a(x-h)^2+k$

Simplify the equation to obtain in standard form equation.

$\begin{gathered} y=-5x^2+x+4 \\ =-5(x^2-\frac{x}{5})+4 \\ =-5(x^2-2\cdot\frac{1}{10}\cdot x+\frac{1}{100}-\frac{1}{100})+4 \\ =-5(x-\frac{1}{10})^2+\frac{1}{20}+4 \\ =-5(x-\frac{1}{10})^2+\frac{81}{20} \end{gathered}$

On compare equation, the vertex is (1/10,81/20).

The axis of symmetry for the standard function is x = h. So axis of symmetry for the function is,

$x=\frac{1}{10}$

Answer:

Vertex: (1/10,81/20) or (0.1,4.05)

Axis of symmetry: x = 1/10 or x = 0.1