To find the [tex]\(x\)[/tex]-intercepts of the quadratic function [tex]\(f(x) = x^2 + 2x - 15\)[/tex], we need to solve for [tex]\(x\)[/tex] when [tex]\(f(x) = 0\)[/tex]. This means we need to solve the equation [tex]\(x^2 + 2x - 15 = 0\)[/tex].
The solutions to the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] are found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the function [tex]\(f(x)\)[/tex], the coefficients are:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 2 \][/tex]
[tex]\[ c = -15 \][/tex]
First, we calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ = 2^2 - 4 \cdot 1 \cdot (-15) \][/tex]
[tex]\[ = 4 + 60 \][/tex]
[tex]\[ = 64 \][/tex]
With the discriminant calculated, we can now find the two solutions for [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{-b + \sqrt{\text{Discriminant}}}{2a} \][/tex]
[tex]\[ x_1 = \frac{-2 + \sqrt{64}}{2 \cdot 1} \][/tex]
[tex]\[ x_1 = \frac{-2 + 8}{2} \][/tex]
[tex]\[ x_1 = \frac{6}{2} \][/tex]
[tex]\[ x_1 = 3 \][/tex]
[tex]\[ x_2 = \frac{-b - \sqrt{\text{Discriminant}}}{2a} \][/tex]
[tex]\[ x_2 = \frac{-2 - \sqrt{64}}{2 \cdot 1} \][/tex]
[tex]\[ x_2 = \frac{-2 - 8}{2} \][/tex]
[tex]\[ x_2 = \frac{-10}{2} \][/tex]
[tex]\[ x_2 = -5 \][/tex]
So, the [tex]\(x\)[/tex]-intercepts of the function [tex]\(f(x)\)[/tex] are [tex]\((3, 0)\)[/tex] and [tex]\((-5, 0)\)[/tex].
To answer the question about the left-most and right-most [tex]\(x\)[/tex]-intercepts, we compare the [tex]\(x\)[/tex]-values:
Left-most [tex]\(x\)[/tex]-intercept: [tex]\((-5, 0)\)[/tex]
Right-most [tex]\(x\)[/tex]-intercept: [tex]\((3, 0)\)[/tex]
