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

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 $- \frac{b}{2 a}$, so in this case it would be
$\frac{- \left(- 4\right)}{2 \cdot 2} = \frac{4}{4} = 1$

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

To find the intercepts it would be best to use quadratic formula:
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$\frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot 2 \cdot 7}}{2 \cdot 2}$
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.