Question

How do you find the vertex and the intercepts for `y= 2x^2 - 4x + 7`?

Answer 1

The vertex is (1,5) and there are no real intercepts.

Explanation:

The x-coordinate of the vertex can be found using `-b/(2a)`, so in this case it would be
`(-(-4))/(2*2) = 4/4 = 1`

So 1 would be the x-coordinate of the vertex. You would then plug this in for x in the original equation:
`y=2(1)^2-4(1)+7` , so `y=5` and the vertex would be (1,5).

To find the intercepts it would be best to use quadratic formula:
`(-b+-sqrt(b^2-4ac))/(2a)`
`(-(-4)+-sqrt((-4)^2-4*2*7))/(2*2)`
When we solve under the square root we can already see that we will get `sqrt(-48)`, which is an imaginary number. When there is an imaginary number under the square root there are no real intercepts, which on the coordinate plane would be a parabola that never touches the x-axis because it either curves back up or back down before it meets the x-axis.