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Basile [38]
3 years ago
15

PLEASE HELP ASAP

Mathematics
1 answer:
Crank3 years ago
7 0

Option D: $f(x)=-2(x+2)(x-4)$ is the function

Explanation:

Let the general form of quadratic equation be $y=a x^{2}+b x+c$

The function passes through the intercepts (-2,0) and $(4,0)$ and also passes though the point $(-1,10)$

Substituting the points (-2,0), $(4,0)$ and $(-1,10)$ in the equation $y=a x^{2}+b x+c$, we get,

4a -2b+c=0  -----------(1)

16a +4b+c=0 ----------(2)

a-b+c=10    -----------(3)

Subtracting (1) and (2), we get,

\ \ 4a -2b+c=0\\ 16a +4b+c=0\\\---------\\ \  \ -12a-6b=0  -----------(4)

Subtracting (2) and (3), we get,

16a +4b+c=0\\\ \ a \ \ - \ \ b \ +c=10\\---------\\15a+5b=-10  ------------(5)

Multiplying equation (4) by 5 and equation (5) by 4, to cancel the term b when adding, we get,

-60a-30b=0\\\ 90a+30b=-60\\--------\\30a=-60\\\ \ a=-2

Thus, the value of a is a=-2

Substituting a=-2 in equation (4), we get,

$\begin{aligned}-12 a+6 b &=0 \\-12(-2)-6 b &=0 \\ 24-6 b &=0 \\ 24 &=6 b \\ 4 &=b \end{aligned}$

Thus, the value of b is b=4

Now, substituting the value of a and b in equation (1), we have,

$\begin{aligned} 4 a-2 b+c &=0 \\ 4(-2)-2(4)+c &=0 \\-8-8+c &=0 \\ c &=16 \end{aligned}$

Thus, the value of c is c=16

Now, substituting the value of a,b and c in the general formula $y=a x^{2}+b x+c$, we get,

y=-2x^{2} +4x+16

Taking out the common term as -2 we get,

y=-2(x^{2} -2x-8)

Factoring , we get,

$y=-2(x+2)(x-4)$

Thus, the function is $f(x)=-2(x+2)(x-4)$

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