To find the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] intercepts of the graph of the equation [tex]\( y = x^2 - 8x + 15 \)[/tex], we'll proceed as follows:
### Finding the [tex]\(y\)[/tex]-Intercept:
The [tex]\(y\)[/tex]-intercept occurs where the graph crosses the [tex]\(y\)[/tex]-axis. This is the point where [tex]\(x = 0\)[/tex].
1. Substitute [tex]\(x = 0\)[/tex] into the equation [tex]\( y = x^2 - 8x + 15 \)[/tex]:
[tex]\[
y = 0^2 - 8 \cdot 0 + 15
\][/tex]
2. Simplifying the expression we get:
[tex]\[
y = 15
\][/tex]
Therefore, the [tex]\(y\)[/tex]-intercept is at the point [tex]\((0, 15)\)[/tex].
### Finding the [tex]\(x\)[/tex]-Intercepts:
The [tex]\(x\)[/tex]-intercepts occur where the graph crosses the [tex]\(x\)[/tex]-axis. These are the points where [tex]\(y = 0\)[/tex].
1. Set [tex]\(y = 0\)[/tex] in the equation [tex]\( y = x^2 - 8x + 15 \)[/tex]:
[tex]\[
0 = x^2 - 8x + 15
\][/tex]
2. To solve for [tex]\(x\)[/tex], we need to factor the quadratic equation. We look for two numbers that multiply to 15 and add up to -8. These numbers are -3 and -5.
3. Factor the quadratic expression:
[tex]\[
(x - 3)(x - 5) = 0
\][/tex]
4. Set each factor equal to zero:
[tex]\[
x - 3 = 0 \quad \text{or} \quad x - 5 = 0
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = 3 \quad \text{or} \quad x = 5
\][/tex]
Therefore, the [tex]\(x\)[/tex]-intercepts are at the points [tex]\((3, 0)\)[/tex] and [tex]\((5, 0)\)[/tex].
### Summary
- The [tex]\(y\)[/tex]-intercept is at [tex]\((0, 15)\)[/tex].
- The [tex]\(x\)[/tex]-intercepts are at [tex]\((3, 0)\)[/tex] and [tex]\((5, 0)\)[/tex].
