Question

How do you find the axis of symmetry, vertex and x intercepts for `y=-x^2-3x`?

Answer 1

Given: `y(x) = ax^2 + bx + c`
axis of symmetry: `x = -b/(2a)`
vertex: `(-b/(2a), y(-b/(2a)))`

Explanation:

Using the general form (above), the axis of symmetry for the given equation, `y(x) = -x^2 - 3x`, is:

`x = -(-3)/(2(-1)`

`x = -3/2`

This is, also, the x coordinate of the vertex.

To find the y coordinate of the vertex, evaluate the function at the x coordinate:

`y(-3/2) = -(-3/2)^2 - 3(-3/2)`

`y(-3/2) = -9/4 + 9/2`

`y(-3/2) = 9/4`

The vertex is:

`(-3/2, 9/4)`

To find the x intercepts, set y(x) = 0:

`0 = -x^2 - 3x`

Multiply both sides by -1:

`0 = x^2 + 3x`

factor:

`0 = x(x + 3)`

`x = 0 and x = -3`

The function intercepts the x axis at the points `(0,0) and (-3, 0)`