Answer :
To find the [tex]\( x \)[/tex]-intercepts of the parabola with the given properties, we will follow a systematic approach. The vertex of the parabola is [tex]\((-2, -8)\)[/tex] and the [tex]\( y \)[/tex]-intercept is [tex]\((0, 4)\)[/tex].
1. Write the equation of the parabola in vertex form:
The vertex form of a parabola is:
[tex]\[ y = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex. For the given vertex [tex]\((-2, -8)\)[/tex], we have:
[tex]\[ y = a(x + 2)^2 - 8 \][/tex]
2. Substitute the [tex]\( y \)[/tex]-intercept to find the value of [tex]\( a \)[/tex]:
The parabola passes through the [tex]\( y \)[/tex]-intercept [tex]\((0, 4)\)[/tex]. Substitute [tex]\((0, 4)\)[/tex] into the equation:
[tex]\[ 4 = a(0 + 2)^2 - 8 \][/tex]
Simplify to solve for [tex]\( a \)[/tex]:
[tex]\[ 4 = a \cdot 4 - 8 \][/tex]
[tex]\[ 4 = 4a - 8 \][/tex]
[tex]\[ 4a = 12 \][/tex]
[tex]\[ a = 3 \][/tex]
3. Write the complete equation with [tex]\( a \)[/tex] determined:
Using [tex]\( a = 3 \)[/tex]:
[tex]\[ y = 3(x + 2)^2 - 8 \][/tex]
4. Set the equation equal to zero to find the [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 3(x + 2)^2 - 8 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 3(x + 2)^2 = 8 \][/tex]
[tex]\[ (x + 2)^2 = \frac{8}{3} \][/tex]
[tex]\[ x + 2 = \pm \sqrt{\frac{8}{3}} \][/tex]
[tex]\[ x = -2 \pm \sqrt{\frac{8}{3}} \][/tex]
5. Calculate the numerical values of the [tex]\( x \)[/tex]-intercepts:
[tex]\[ x_1 = -2 + \sqrt{\frac{8}{3}} \][/tex]
[tex]\[ x_2 = -2 - \sqrt{\frac{8}{3}} \][/tex]
Using approximate calculations, we get:
[tex]\[ \sqrt{\frac{8}{3}} \approx 1.63299316185545 \][/tex]
Therefore:
[tex]\[ x_1 = -2 + 1.63299316185545 \approx -0.367006838144548 \][/tex]
[tex]\[ x_2 = -2 - 1.63299316185545 \approx -3.63299316185545 \][/tex]
6. Write the [tex]\( x \)[/tex]-intercepts as coordinates:
[tex]\[ \left(x_1, 0\right), \left(x_2, 0\right) \Rightarrow (-0.37, 0), (-3.63, 0) \][/tex]
Therefore, the [tex]\( x \)[/tex]-intercepts of the parabola are [tex]\( \left(-3.63, 0\right) \)[/tex] and [tex]\( \left(-0.37, 0\right) \)[/tex].
1. Write the equation of the parabola in vertex form:
The vertex form of a parabola is:
[tex]\[ y = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex. For the given vertex [tex]\((-2, -8)\)[/tex], we have:
[tex]\[ y = a(x + 2)^2 - 8 \][/tex]
2. Substitute the [tex]\( y \)[/tex]-intercept to find the value of [tex]\( a \)[/tex]:
The parabola passes through the [tex]\( y \)[/tex]-intercept [tex]\((0, 4)\)[/tex]. Substitute [tex]\((0, 4)\)[/tex] into the equation:
[tex]\[ 4 = a(0 + 2)^2 - 8 \][/tex]
Simplify to solve for [tex]\( a \)[/tex]:
[tex]\[ 4 = a \cdot 4 - 8 \][/tex]
[tex]\[ 4 = 4a - 8 \][/tex]
[tex]\[ 4a = 12 \][/tex]
[tex]\[ a = 3 \][/tex]
3. Write the complete equation with [tex]\( a \)[/tex] determined:
Using [tex]\( a = 3 \)[/tex]:
[tex]\[ y = 3(x + 2)^2 - 8 \][/tex]
4. Set the equation equal to zero to find the [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 3(x + 2)^2 - 8 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 3(x + 2)^2 = 8 \][/tex]
[tex]\[ (x + 2)^2 = \frac{8}{3} \][/tex]
[tex]\[ x + 2 = \pm \sqrt{\frac{8}{3}} \][/tex]
[tex]\[ x = -2 \pm \sqrt{\frac{8}{3}} \][/tex]
5. Calculate the numerical values of the [tex]\( x \)[/tex]-intercepts:
[tex]\[ x_1 = -2 + \sqrt{\frac{8}{3}} \][/tex]
[tex]\[ x_2 = -2 - \sqrt{\frac{8}{3}} \][/tex]
Using approximate calculations, we get:
[tex]\[ \sqrt{\frac{8}{3}} \approx 1.63299316185545 \][/tex]
Therefore:
[tex]\[ x_1 = -2 + 1.63299316185545 \approx -0.367006838144548 \][/tex]
[tex]\[ x_2 = -2 - 1.63299316185545 \approx -3.63299316185545 \][/tex]
6. Write the [tex]\( x \)[/tex]-intercepts as coordinates:
[tex]\[ \left(x_1, 0\right), \left(x_2, 0\right) \Rightarrow (-0.37, 0), (-3.63, 0) \][/tex]
Therefore, the [tex]\( x \)[/tex]-intercepts of the parabola are [tex]\( \left(-3.63, 0\right) \)[/tex] and [tex]\( \left(-0.37, 0\right) \)[/tex].
