Question

What is the axis of symmetry and vertex for the graph `y=2x^2 +16x - 12`?

Answer 1

Axis of symmetry is `x=-4`

Vertex is `(-4,-44)`

Explanation:

In a quadratic equation `f(x)=ax^2+bx+c` you can find the axis of symmetry by using the equation `-b/(2a)`

You can find the vertex with this formula: `(-b/(2a),f(-b/(2a)))`

In the question, `a=2, b=16, c=-12`

So the axis of symmetry can be found by evaluating:

`-16/(2(2))=-16/4=-4`

To find the vertex, we use the axis of symmetry as the x-coordinate and plug in the x-value into the function for the y-coordinate:

`f(-4)=2(-4)^2+16(-4)-12`
`f(-4)=2*16-64-12`
`f(-4)=32-64-12`
`f(-4)=-32-12`
`f(-4)=-44`

Thus the vertex is `(-4,-44)`