# How do you find the vertex and the intercepts for y=-2x^2 + 4x - 3?

Nov 13, 2017

Vertex is (1, -1)
There are no real x-intercepts.
Y-intercepts are at (0, -3)

#### Explanation:

First, using $a {x}^{2} + b x + c$ as a model, let's get the values of $a$, $b$, and $c$ for this polynomial.

$a = - 2$
$b = 4$
$c = - 3$

The x-coordinate of the vertex is found by calculating $- \frac{b}{2 a}$,

$- \frac{4}{2 \cdot - 2} = 1$

Now, we plug in $x = 1$ into our equation to get it's y-coordinate.

$y = - 2 {\left(1\right)}^{2} + 4 \left(1\right) - 3$
$y = - 1$

So we can reasonably conclude that the vertex is the point (1, -1). In order to find the x-intercepts, it is the values of $x$ of the equation when $y = 0$.

I'll save you the time, the polynomial does not factor easy. So we can then use the quadratic equation to calculate the values of $x$ when $y = 0$. However, let's first look at the discriminant, which reveals what types of roots the equation has.

${b}^{2} - 4 a c$
${4}^{2} - 4 \cdot \left(- 2\right) \cdot \left(- 3\right)$
$16 - 24 = - 8$

Since the discriminant is negative, it's context in the quadratic formula is $\sqrt{- 8}$ which is imaginary. This means that the two roots (indicated by the power of the polynomial), are a pair of imaginary roots. Consequently, we can reasonably conclude that the polynomial has no real roots, or no real x-intercepts.

On another note, let's go to the y-intercept(s). We can say that the graph intercepts the y-axis where $x = 0$, so we can just plug $x = 0$ into the polynomial.

$y = - 2 {\left(0\right)}^{2} + 4 \left(0\right) - 3$
$y = - 3$

With this, we can say (0, -3) is a point and that it is the only y-intercept of the equation.