Find the distance between the pair of points: (-6, 4) and (6,4)
1 answer:
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9514 1404 393
Answer:
y = 1/2x^2 +x -15/2
Step-by-step explanation:
The vertex is the point where the graph reaches is extreme: (-1, -8). The y-intercept is (0, -7.5). The x-intercepts are (-5, 0) and (3, 0).
When the quadratic is written in the form ...
y = x^2 +bx +c
the value of b is the opposite of the sum of the x-intercepts. The value of c is their product. These facts would give ...
y = x^2 -(-5 +3)x +(-5)(3) = x^2 +2x -15
The constant in this equation is -15, which is the value y has when x=0. Our actual y-intercept is half that, so the graph we see is this equation scaled by a factor of 1/2:
y = 1/2x^2 +x -15/2
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<em>Alternate solution</em>
You can start with the vertex form.
y = a(x -h)^2 +k . . . . . . . vertical scale factor 'a' and vertex (h, k)
On the graph, the vertical scale factor is the vertical difference between the low point (vertex) and the point that is 1 unit right or left. Here, that is ...
-7.5 - (-8) = 0.5 = a
We've already said the vertex is (h, k) = (-1, -8), so the vertex form equation is ...
y = 0.5(x +1)^2 -8
We know the square of a binomial is (a+b)^2 = a^2 +2ab +b^2, so we have ...
y = 0.5(x^2 +2x +1) -8
y = 0.5x^2 +x +0.5 -8 . . . . . eliminate parentheses
y = 0.5x^2 +x -7.5
As far as I can tell, that is as far as the expression can be simplified :)