How do you find the vertex and the intercepts for y=2x2−4x+7?

1 Answer
Sarah D.
Oct 18, 2017

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

Explanation:

The x-coordinate of the vertex can be found using −b2a, so in this case it would be
−(−4)2⋅2=44=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±√b2−4ac2a
−(−4)±√(−4)2−4⋅2⋅72⋅2
When we solve under the square root we can already see that we will get √−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.

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