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Sagot :
Answer:
x-intercepts
[tex](- 3, 0) $ and $(5, 0)[/tex]
Vertex is at [tex](1, -16)[/tex]
Step-by-step explanation:
The given parabola equation is obviously incorrect since solving it will result in non-real roots
Taking the equation as
[tex]y = x^2 - 2x - 15[/tex]
x-intercepts of a parabola are the points where the parabola intercepts the x-axis
Equivalently these are also the points where the y-value is 0
Given the parabola
[tex]x^2 - 2x - 15[/tex] we can find the x intercepts by setting this equation to 0 and solving for x
[tex]Solve \;x^2 - 2x - 15 = 0[/tex]
[tex]x^2 - 2x - 15[/tex] can be factored as [tex](x - 5)(x +3)[/tex]
Therefore
(x - 5)(x + 3) = 0
==> x - 5 = 0 or x = 5
==> x + 3 = 0 or x = -3
- The x-intercepts are at (-3, 0) and (5, 0)
Vertex can be obtained by taking the first derivative of the equation, setting to 0 and solving for x and then solving for y
First derivative of [tex]x^2 - 2x - 15 = 2x - 2\\[/tex]
2x -2 = 0 ==> x = 1
Plug x = 1 into the equation [tex]y = x^2 - 2x - 15[/tex]:
y = 1^2 - 2(1) - 15
y = 1 - 2 -15 = - 16
- Vertex is at (1, -16)
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