Question

How do you find the vertex and axis of symmetry, and then graph the parabola given by: `f(x)=2x^2-6x-5`?

Answer 1

The x-coordinate of the vertex is equal to `-b/(2a)`

Explanation:

The general format of a quadratic equation is:

`f(x) = ax^2+bx+c`

So, for this problem:

`a = 2`
`b=-6`
`c=-5`

The vertex (x,y) has an x value:
`x = -b/(2a) = -(-6)/(2*2) = 1.5`

To find the y value of the vertex, simply plug the x value (1.5) back into the equation:

`y = 2(1.5^2)-6(1.5)-5 = -9.5`

So the vertex is `(1.5, -9.5)`

The axis of symmetry travels vertical through the x value of the vetex:

axis of symmetry: `x = 1.5`

In order to graph the function, simply set up an x-y table and plug some x values {-5,-4,-3,-2,-1,0,1,2,3,4,5} into the equation to solve for y:

graph{2x^2-6x-5 [-18.3, 21.7, -10.8, 9.2]}

Hope that helped