Question

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

Answer 1

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

Explanation:

First, using `ax^2 + bx + 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 `-b/(2a)`,

`-(4)/(2*-2) = 1`

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

`y = -2(1)^2 + 4(1) - 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 - 4ac`
`4^2 - 4 * (-2) * (-3)`
`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(0)^2 + 4(0) - 3`
`y = -3`

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