Option D: [tex]$f(x)=-2(x+2)(x-4)$[/tex] is the function
Explanation:
Let the general form of quadratic equation be [tex]$y=a x^{2}+b x+c$[/tex]
The function passes through the intercepts [tex](-2,0)[/tex] and [tex]$(4,0)$[/tex] and also passes though the point [tex]$(-1,10)$[/tex]
Substituting the points [tex](-2,0)[/tex], [tex]$(4,0)$[/tex] and [tex]$(-1,10)$[/tex] in the equation [tex]$y=a x^{2}+b x+c$[/tex], we get,
[tex]4a -2b+c=0[/tex] -----------(1)
[tex]16a +4b+c=0[/tex] ----------(2)
[tex]a-b+c=10[/tex] -----------(3)
Subtracting (1) and (2), we get,
[tex]\ \ 4a -2b+c=0\\ 16a +4b+c=0\\\---------\\ \ \ -12a-6b=0[/tex] -----------(4)
Subtracting (2) and (3), we get,
[tex]16a +4b+c=0\\\ \ a \ \ - \ \ b \ +c=10\\---------\\15a+5b=-10[/tex] ------------(5)
Multiplying equation (4) by 5 and equation (5) by 4, to cancel the term b when adding, we get,
[tex]-60a-30b=0\\\ 90a+30b=-60\\--------\\30a=-60\\\ \ a=-2[/tex]
Thus, the value of a is [tex]a=-2[/tex]
Substituting [tex]a=-2[/tex] in equation (4), we get,
[tex]$\begin{aligned}-12 a+6 b &=0 \\-12(-2)-6 b &=0 \\ 24-6 b &=0 \\ 24 &=6 b \\ 4 &=b \end{aligned}$[/tex]
Thus, the value of b is [tex]b=4[/tex]
Now, substituting the value of a and b in equation (1), we have,
[tex]$\begin{aligned} 4 a-2 b+c &=0 \\ 4(-2)-2(4)+c &=0 \\-8-8+c &=0 \\ c &=16 \end{aligned}$[/tex]
Thus, the value of c is [tex]c=16[/tex]
Now, substituting the value of a,b and c in the general formula [tex]$y=a x^{2}+b x+c$[/tex], we get,
[tex]y=-2x^{2} +4x+16[/tex]
Taking out the common term as -2 we get,
[tex]y=-2(x^{2} -2x-8)[/tex]
Factoring , we get,
[tex]$y=-2(x+2)(x-4)$[/tex]
Thus, the function is [tex]$f(x)=-2(x+2)(x-4)$[/tex]
