To determine the [tex]\(x\)[/tex]-intercepts of the quadratic equation [tex]\(y = x^2 + 7x + 12\)[/tex], we need to solve for the values of [tex]\(x\)[/tex] where [tex]\(y = 0\)[/tex]. This translates to solving the quadratic equation [tex]\(x^2 + 7x + 12 = 0\)[/tex].
To solve this quadratic equation, we'll use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients from the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex]. Here, [tex]\(a = 1\)[/tex], [tex]\(b = 7\)[/tex], and [tex]\(c = 12\)[/tex].
First, we calculate the discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac = 7^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1
\][/tex]
Now we use the quadratic formula to find the two solutions for [tex]\(x\)[/tex]:
For the positive square root:
[tex]\[
x_1 = \frac{-b + \sqrt{\text{Discriminant}}}{2a} = \frac{-7 + \sqrt{1}}{2 \cdot 1} = \frac{-7 + 1}{2} = \frac{-6}{2} = -3
\][/tex]
For the negative square root:
[tex]\[
x_2 = \frac{-b - \sqrt{\text{Discriminant}}}{2a} = \frac{-7 - \sqrt{1}}{2 \cdot 1} = \frac{-7 - 1}{2} = \frac{-8}{2} = -4
\][/tex]
Thus, the [tex]\(x\)[/tex]-intercepts of the graph of [tex]\(y = x^2 + 7x + 12\)[/tex] are at [tex]\((-3, 0)\)[/tex] and [tex]\((-4, 0)\)[/tex].
Therefore, the correct answer is:
B. [tex]\((-4,0)\)[/tex] and [tex]\((-3,0)\)[/tex]
