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Sagot :
Sure! Let's go through the solution step-by-step for the quadratic function [tex]\( f(x) = x^2 - 4x - 12 \)[/tex].
### a) Determine whether the graph of the given quadratic function opens up or down
We need to look at the coefficient of the [tex]\( x^2 \)[/tex] term. If the coefficient (which is [tex]\( a \)[/tex]) is positive, the parabola opens up. If it is negative, the parabola opens down.
For [tex]\( f(x) = x^2 - 4x - 12 \)[/tex], the coefficient of [tex]\( x^2 \)[/tex] is [tex]\( 1 \)[/tex], which is positive.
#### Answer:
The parabola opens up.
### b) Find the vertex
The vertex of a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula for [tex]\( h \)[/tex] and [tex]\( k \)[/tex]:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) \][/tex]
For [tex]\( f(x) = x^2 - 4x - 12 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -4 \)[/tex]
So,
[tex]\[ h = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2 \][/tex]
Now find [tex]\( k = f(h) \)[/tex]:
[tex]\[ k = f(2) = (2)^2 - 4(2) - 12 \][/tex]
[tex]\[ k = 4 - 8 - 12 \][/tex]
[tex]\[ k = -16 \][/tex]
#### Answer:
The vertex of the parabola is [tex]\((2, -16)\)[/tex].
### c) Find the axis of symmetry
The axis of symmetry is a vertical line that passes through the vertex. Therefore, it is [tex]\( x = h \)[/tex].
#### Answer:
The axis of symmetry is [tex]\( x = 2 \)[/tex].
### d) Find the [tex]\( x \)[/tex]-intercepts and [tex]\( y \)[/tex]-intercepts
#### [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x^2 - 4x - 12 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{4 \pm \sqrt{16 + 48}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 8}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{4 + 8}{2} = 6 \][/tex]
[tex]\[ x = \frac{4 - 8}{2} = -2 \][/tex]
#### Answer:
The [tex]\( x \)[/tex]-intercepts are [tex]\(-2\)[/tex] and [tex]\(6\)[/tex].
#### [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 4 \cdot 0 - 12 \][/tex]
[tex]\[ f(0) = -12 \][/tex]
#### Answer:
The [tex]\( y \)[/tex]-intercept is [tex]\(-12\)[/tex].
### e) Sketch the graph of the function
With the information gathered:
- The parabola opens up.
- The vertex is at [tex]\((2, -16)\)[/tex].
- The axis of symmetry is the line [tex]\( x = 2 \)[/tex].
- The [tex]\( x \)[/tex]-intercepts are at [tex]\(-2\)[/tex] and [tex]\(6\)[/tex].
- The [tex]\( y \)[/tex]-intercept is at [tex]\(-12\)[/tex].
You can use these key points and symmetry to sketch the parabola. The vertex is the lowest point since it opens up, and the graph will pass through the intercepts identified.
Here's a sketch of the parabola:
1. Plot the vertex [tex]\((2, -16)\)[/tex].
2. Draw the axis of symmetry [tex]\( x = 2 \)[/tex].
3. Plot the [tex]\( x \)[/tex]-intercepts at [tex]\(-2, 0 \)[/tex] and [tex]\(6, 0\)[/tex].
4. Plot the [tex]\( y \)[/tex]-intercept at [tex]\((0, -12)\)[/tex].
5. Draw the curve of the parabola opening upwards, passing through these points.
This completes the detailed analysis and sketch of the quadratic function [tex]\( f(x) = x^2 - 4x - 12 \)[/tex].
### a) Determine whether the graph of the given quadratic function opens up or down
We need to look at the coefficient of the [tex]\( x^2 \)[/tex] term. If the coefficient (which is [tex]\( a \)[/tex]) is positive, the parabola opens up. If it is negative, the parabola opens down.
For [tex]\( f(x) = x^2 - 4x - 12 \)[/tex], the coefficient of [tex]\( x^2 \)[/tex] is [tex]\( 1 \)[/tex], which is positive.
#### Answer:
The parabola opens up.
### b) Find the vertex
The vertex of a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula for [tex]\( h \)[/tex] and [tex]\( k \)[/tex]:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) \][/tex]
For [tex]\( f(x) = x^2 - 4x - 12 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -4 \)[/tex]
So,
[tex]\[ h = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2 \][/tex]
Now find [tex]\( k = f(h) \)[/tex]:
[tex]\[ k = f(2) = (2)^2 - 4(2) - 12 \][/tex]
[tex]\[ k = 4 - 8 - 12 \][/tex]
[tex]\[ k = -16 \][/tex]
#### Answer:
The vertex of the parabola is [tex]\((2, -16)\)[/tex].
### c) Find the axis of symmetry
The axis of symmetry is a vertical line that passes through the vertex. Therefore, it is [tex]\( x = h \)[/tex].
#### Answer:
The axis of symmetry is [tex]\( x = 2 \)[/tex].
### d) Find the [tex]\( x \)[/tex]-intercepts and [tex]\( y \)[/tex]-intercepts
#### [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x^2 - 4x - 12 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{4 \pm \sqrt{16 + 48}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{64}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 8}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{4 + 8}{2} = 6 \][/tex]
[tex]\[ x = \frac{4 - 8}{2} = -2 \][/tex]
#### Answer:
The [tex]\( x \)[/tex]-intercepts are [tex]\(-2\)[/tex] and [tex]\(6\)[/tex].
#### [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 4 \cdot 0 - 12 \][/tex]
[tex]\[ f(0) = -12 \][/tex]
#### Answer:
The [tex]\( y \)[/tex]-intercept is [tex]\(-12\)[/tex].
### e) Sketch the graph of the function
With the information gathered:
- The parabola opens up.
- The vertex is at [tex]\((2, -16)\)[/tex].
- The axis of symmetry is the line [tex]\( x = 2 \)[/tex].
- The [tex]\( x \)[/tex]-intercepts are at [tex]\(-2\)[/tex] and [tex]\(6\)[/tex].
- The [tex]\( y \)[/tex]-intercept is at [tex]\(-12\)[/tex].
You can use these key points and symmetry to sketch the parabola. The vertex is the lowest point since it opens up, and the graph will pass through the intercepts identified.
Here's a sketch of the parabola:
1. Plot the vertex [tex]\((2, -16)\)[/tex].
2. Draw the axis of symmetry [tex]\( x = 2 \)[/tex].
3. Plot the [tex]\( x \)[/tex]-intercepts at [tex]\(-2, 0 \)[/tex] and [tex]\(6, 0\)[/tex].
4. Plot the [tex]\( y \)[/tex]-intercept at [tex]\((0, -12)\)[/tex].
5. Draw the curve of the parabola opening upwards, passing through these points.
This completes the detailed analysis and sketch of the quadratic function [tex]\( f(x) = x^2 - 4x - 12 \)[/tex].
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