Question

please helpp (15 points)

Answer 1

We have:

• vertex = (-1, 16)
• vertical intercept =  (0, 14)
• x-intercepts = (-3,83, 0), (.1,83, 0).

How to find the vertex of the parabola?

For a parabola like:

$y = a*x^2 + b*x + c$

The vertex is at:

$x = -b/2a$

In this case, we have:

$y = -2x^2 -4x + 14$

So the vertex is at:

$x = 4/2*(-2) = -1$

To get the y-value, we need to evaluate in x = -1.

$y = -2*(-1)^2 - 4*(-1) + 14 = -2 + 4 + 14 = 16$

So the vertex is (-1, 16)

How to find the y-intercept?

To do that we just evaluate in x = 0.

$y = -2*0^2 -4*0 + 14 = 14$

The vertical intercept is (0, 14).

How to get the x-intercepts?

Using Bhaskara's formula we get:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4*(-2)*14} }{2*-2} \\\\x = \frac{4 \pm 11.31 }{-4}$

So the two roots are:

x = (4 + 11.31)/(-4) = -3.83

x = (4 - 11.31)/(-4) = 1.83

If you want to learn more about parabolas:

https://brainly.com/question/4061870

#SPJ1