Answer :
Certainly! Let's go through the problem step-by-step for the quadratic equation [tex]\(y = x^2 + 6x - 16\)[/tex]:
### (a) Determine whether the parabola opens upward or downward.
For a quadratic equation in the form [tex]\(y = ax^2 + bx + c\)[/tex], the direction in which the parabola opens is determined by the coefficient [tex]\(a\)[/tex]:
- If [tex]\(a > 0\)[/tex], the parabola opens upward.
- If [tex]\(a < 0\)[/tex], the parabola opens downward.
In this equation, [tex]\(a = 1\)[/tex], which is greater than 0. Therefore, the parabola opens upward.
### (b) Find the vertex.
The vertex of a parabola [tex]\(y = ax^2 + bx + c\)[/tex] can be found using the formula:
- The [tex]\(x\)[/tex]-coordinate of the vertex, [tex]\(h\)[/tex], is given by [tex]\(h = -\frac{b}{2a}\)[/tex].
- The [tex]\(y\)[/tex]-coordinate of the vertex, [tex]\(k\)[/tex], can be found by substituting [tex]\(h\)[/tex] back into the equation.
For this equation, [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -16\)[/tex]:
- [tex]\(h = -\frac{b}{2a} = -\frac{6}{2 \times 1} = -3\)[/tex]
- To find [tex]\(k\)[/tex], substitute [tex]\(h = -3\)[/tex] back into the equation:
[tex]\[ k = (1)(-3)^2 + 6(-3) - 16 = 9 - 18 - 16 = -25 \][/tex]
So, the vertex is [tex]\((-3, -25)\)[/tex].
### (c) Find the [tex]\(x\)[/tex]-intercept(s), if possible.
The [tex]\(x\)[/tex]-intercepts are the points where [tex]\(y = 0\)[/tex]. For the quadratic equation [tex]\(y = ax^2 + bx + c\)[/tex]:
- Set [tex]\(ax^2 + bx + c = 0\)[/tex] and solve for [tex]\(x\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For this equation, [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -16\)[/tex]:
- Calculate the discriminant, [tex]\(\Delta = b^2 - 4ac = 6^2 - 4(1)(-16) = 36 + 64 = 100\)[/tex], which is positive.
- Since the discriminant is positive, there are two real solutions:
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2 \times 1} = \frac{-6 \pm 10}{2} \][/tex]
This simplifies to:
[tex]\[ x_1 = \frac{-6 + 10}{2} = 2, \quad x_2 = \frac{-6 - 10}{2} = -8 \][/tex]
So, the [tex]\(x\)[/tex]-intercepts are [tex]\(x = 2\)[/tex] and [tex]\(x = -8\)[/tex].
### (d) Find the [tex]\(y\)[/tex]-intercept.
The [tex]\(y\)[/tex]-intercept occurs where [tex]\(x = 0\)[/tex]:
- Substitute [tex]\(x = 0\)[/tex] into the equation [tex]\(y = x^2 + 6x - 16\)[/tex]:
[tex]\[ y = (0)^2 + 6(0) - 16 = -16 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is [tex]\(-16\)[/tex].
### (e) Sketch the graph.
Using the information we've gathered:
- The parabola opens upward.
- The vertex is at [tex]\((-3, -25)\)[/tex].
- The [tex]\(x\)[/tex]-intercepts are at [tex]\(x = 2\)[/tex] and [tex]\(x = -8\)[/tex].
- The [tex]\(y\)[/tex]-intercept is at [tex]\(-16\)[/tex].
A rough sketch of the graph would show:
- A parabola opening upward.
- A vertex at [tex]\((-3, -25)\)[/tex].
- Points crossing the [tex]\(x\)[/tex]-axis at [tex]\(2\)[/tex] and [tex]\(-8\)[/tex].
- A point crossing the [tex]\(y\)[/tex]-axis at [tex]\(-16\)[/tex].
This comprehensive analysis helps us understand the behavior and key features of the quadratic function [tex]\(y = x^2 + 6x - 16\)[/tex].
### (a) Determine whether the parabola opens upward or downward.
For a quadratic equation in the form [tex]\(y = ax^2 + bx + c\)[/tex], the direction in which the parabola opens is determined by the coefficient [tex]\(a\)[/tex]:
- If [tex]\(a > 0\)[/tex], the parabola opens upward.
- If [tex]\(a < 0\)[/tex], the parabola opens downward.
In this equation, [tex]\(a = 1\)[/tex], which is greater than 0. Therefore, the parabola opens upward.
### (b) Find the vertex.
The vertex of a parabola [tex]\(y = ax^2 + bx + c\)[/tex] can be found using the formula:
- The [tex]\(x\)[/tex]-coordinate of the vertex, [tex]\(h\)[/tex], is given by [tex]\(h = -\frac{b}{2a}\)[/tex].
- The [tex]\(y\)[/tex]-coordinate of the vertex, [tex]\(k\)[/tex], can be found by substituting [tex]\(h\)[/tex] back into the equation.
For this equation, [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -16\)[/tex]:
- [tex]\(h = -\frac{b}{2a} = -\frac{6}{2 \times 1} = -3\)[/tex]
- To find [tex]\(k\)[/tex], substitute [tex]\(h = -3\)[/tex] back into the equation:
[tex]\[ k = (1)(-3)^2 + 6(-3) - 16 = 9 - 18 - 16 = -25 \][/tex]
So, the vertex is [tex]\((-3, -25)\)[/tex].
### (c) Find the [tex]\(x\)[/tex]-intercept(s), if possible.
The [tex]\(x\)[/tex]-intercepts are the points where [tex]\(y = 0\)[/tex]. For the quadratic equation [tex]\(y = ax^2 + bx + c\)[/tex]:
- Set [tex]\(ax^2 + bx + c = 0\)[/tex] and solve for [tex]\(x\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For this equation, [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -16\)[/tex]:
- Calculate the discriminant, [tex]\(\Delta = b^2 - 4ac = 6^2 - 4(1)(-16) = 36 + 64 = 100\)[/tex], which is positive.
- Since the discriminant is positive, there are two real solutions:
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2 \times 1} = \frac{-6 \pm 10}{2} \][/tex]
This simplifies to:
[tex]\[ x_1 = \frac{-6 + 10}{2} = 2, \quad x_2 = \frac{-6 - 10}{2} = -8 \][/tex]
So, the [tex]\(x\)[/tex]-intercepts are [tex]\(x = 2\)[/tex] and [tex]\(x = -8\)[/tex].
### (d) Find the [tex]\(y\)[/tex]-intercept.
The [tex]\(y\)[/tex]-intercept occurs where [tex]\(x = 0\)[/tex]:
- Substitute [tex]\(x = 0\)[/tex] into the equation [tex]\(y = x^2 + 6x - 16\)[/tex]:
[tex]\[ y = (0)^2 + 6(0) - 16 = -16 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is [tex]\(-16\)[/tex].
### (e) Sketch the graph.
Using the information we've gathered:
- The parabola opens upward.
- The vertex is at [tex]\((-3, -25)\)[/tex].
- The [tex]\(x\)[/tex]-intercepts are at [tex]\(x = 2\)[/tex] and [tex]\(x = -8\)[/tex].
- The [tex]\(y\)[/tex]-intercept is at [tex]\(-16\)[/tex].
A rough sketch of the graph would show:
- A parabola opening upward.
- A vertex at [tex]\((-3, -25)\)[/tex].
- Points crossing the [tex]\(x\)[/tex]-axis at [tex]\(2\)[/tex] and [tex]\(-8\)[/tex].
- A point crossing the [tex]\(y\)[/tex]-axis at [tex]\(-16\)[/tex].
This comprehensive analysis helps us understand the behavior and key features of the quadratic function [tex]\(y = x^2 + 6x - 16\)[/tex].
