Question 1181909: What is the factored form equation for a parabola that has x-intercepts at -3 and 1 and vertex point at (- 1, -8).
Found 3 solutions by MathLover1, math_tutor2020, josgarithmetic:
Answer by MathLover1(20849)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
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If p and q are roots of a quadratic, then x-p and x-q are the two factors.
Based on that, we have the roots -3 and 1 lead us to the factors (x+3) and (x-1)
Notice that the expression (x+3)(x-1), when set equal to zero, will result in x = -3 or x = 1 as the two roots. I'm using the zero product property.
In other words, solving (x+3)(x-1) = 0 will get us x = -3 and x = 1 as the two roots.
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That takes care of the x intercept portion. Now let's consider the vertex point.
Let's see if plugging x = -1 leads to y = -8
(x+3)(x-1) = (-1+3)(-1-1) = (2)(-2) = -4
Unfortunately, we don't reach the target we want.
We can fix this by sticking a 2 out front, so that we scale the -4 up to -8
Trying x = -1 again gets us
2(x+3)(x-1) = 2(-1+3)(-1-1) = 2(2)(-2) = -8
and now everything works out
Answer: 2(x+3)(x-1)
Graph:
Points A and B are the roots, aka x intercepts.
Point C is the vertex.
Because the vertex is below the x axis, and there are two x intercepts, this must mean the parabola opens upward (hence the positive leading coefficient a = 2)
Answer by josgarithmetic(39617) (Show Source):
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