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:
[tex]y = a*x^2 + b*x + c[/tex]
The vertex is at:
[tex]x = -b/2a[/tex]
In this case, we have:
[tex]y = -2x^2 -4x + 14[/tex]
So the vertex is at:
[tex]x = 4/2*(-2) = -1[/tex]
To get the y-value, we need to evaluate in x = -1.
[tex]y = -2*(-1)^2 - 4*(-1) + 14 = -2 + 4 + 14 = 16[/tex]
So the vertex is (-1, 16)
How to find the y-intercept?
To do that we just evaluate in x = 0.
[tex]y = -2*0^2 -4*0 + 14 = 14[/tex]
The vertical intercept is (0, 14).
How to get the x-intercepts?
Using Bhaskara's formula we get:
[tex]x = \frac{4 \pm \sqrt{(-4)^2 - 4*(-2)*14} }{2*-2} \\\\x = \frac{4 \pm 11.31 }{-4}[/tex]
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
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