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

1 Answer
Oct 18, 2017

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


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:
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.