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3 years ago

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PLEASEEEE HELP IW ILL GIVE BRAINLIest

1 answer:

Temka [501]3 years ago

3 0

Answer:

t1= 2-3-6=-7

t2=8-6-6=-4

t3= 18-9-6=3

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PLEASE HELP ASAP

Crank

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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