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PIT_PIT [208]
3 years ago
7

Find the x-intercepts of the parabola with vertex (-4,2) and y-intercept (0,-30). Write your answer in this form: (x1, y1),(x2,

y2). If necessary, round to the nearest hundredth.
Mathematics
1 answer:
Ainat [17]3 years ago
3 0
Vertex form of a parabola 
<span>y = a (x - h)^2 + k </span>

<span>where (h, k) is the vertex </span>
Substituting the values of h and k.
we get, 

<span>y = a(x + 4)^2 + 2 </span>

<span>substituting in the point (0, -30) for  x and  y
</span><span>-30 = a (0 + 4)^2 + 2 

</span>solve for a,

<span>-30 = 16 a + 2 </span>
<span>-32 = 16 a </span>
<span>-2 = a </span>

<span>y = -2(x + 4)^2 + 2 </span>

<span>Put y = 0 </span>

<span>-2 x^2 - 16 x - 30 = 0 </span>
<span>-2(x^2 + 8 x + 15) = 0 </span>
<span>x^2 + 8 x + 15 = 0 </span>
<span>(x + 3)(x + 5) = 0 </span>

<span>x = -3
x = -5</span>
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Step-by-step explanation:

Given

$\[x^2 + 22x + \underline{~~~~}.\]$

Required

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