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The graph of the quadratic function [tex]f[/tex] has [tex]x[/tex]-intercepts of [tex](-7, 0)[/tex] and [tex](-4, 0)[/tex]. Which equation could represent the function?

A. [tex]f(x)=-\frac{1}{2}(x-7)(x+4)[/tex]

B. [tex]f(x)=2(x+7)(x-4)[/tex]

C. [tex]f(x)=(x-7)(x-4)[/tex]

D. [tex]f(x)=-3(x+7)(x+4)[/tex]


Sagot :

To find the quadratic function [tex]\( f(x) \)[/tex] that has [tex]\( x \)[/tex]-intercepts at [tex]\((-7,0)\)[/tex] and [tex]\((-4,0)\)[/tex], we need to use the roots of the function. Remember that [tex]\( x \)[/tex]-intercepts (or roots) of a quadratic function are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].

A general form for a quadratic equation given roots [tex]\( p \)[/tex] and [tex]\( q \)[/tex] is:
[tex]\[ f(x) = a(x - p)(x - q) \][/tex]

In this case, the [tex]\( x \)[/tex]-intercepts are [tex]\( x = -7 \)[/tex] and [tex]\( x = -4 \)[/tex]. Plugging these into the general form, the equation becomes:
[tex]\[ f(x) = a(x + 7)(x + 4) \][/tex]

We need to determine the correct form that matches this structure from the given choices. Let's analyze each option:

Option A: [tex]\( f(x) = -\frac{1}{2}(x - 7)(x + 4) \)[/tex]
- This converts to [tex]\( f(x) = -\frac{1}{2}(x + 7)(x + 4) \)[/tex]
- The roots would be 7 and -4, which do not match [tex]\(-7\)[/tex] and [tex]\(-4\)[/tex]. Therefore, this is incorrect.

Option B: [tex]\( f(x) = 2(x + 7)(x - 4) \)[/tex]
- This converts to [tex]\( f(x) = 2(x + 7)(x - 4) \)[/tex]
- The roots would be -7 and 4, which do not match [tex]\(-7\)[/tex] and [tex]\(-4\)[/tex]. Therefore, this is incorrect.

Option C: [tex]\( f(x) = (x - 7)(x - 4) \)[/tex]
- This converts to [tex]\( f(x) = (x - 7)(x - 4) \)[/tex]
- The roots would be 7 and 4, which do not match [tex]\(-7\)[/tex] and [tex]\(-4\)[/tex]. Therefore, this is incorrect.

Option D: [tex]\( f(x) = -3(x + 7)(x + 4) \)[/tex]
- This converts to [tex]\( f(x) = -3(x + 7)(x + 4) \)[/tex]
- The roots are indeed -7 and -4, which matches our given [tex]\( x \)[/tex]-intercepts.

Thus, the correct answer is:
[tex]\[ \boxed{D} \][/tex]