Math / Algebra

QuestionUse the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the parabola's axis graph to determine the domain and range of the function. $f(x)=4 x-x^{2}+12$
Use the graphing tool to graph the equation. Use the vertex and one of the intercepts to draw the graph.
The axis of symmetry is $\square$ (Type an equation.) The domain of the function is $\square$ (Type your answer in interval notation.) The range of the function is $\square$ . (Type your answer in interval notation.)

Studdy Solution

STEP 1

1. The function is a quadratic function of the form $f(x) = ax^2 + bx + c$ .
2. The vertex form of a quadratic function can be used to find the vertex.
3. The axis of symmetry can be found using the formula $x = -\frac{b}{2a}$ .
4. The domain of a quadratic function is all real numbers.
5. The range of the function can be determined by the vertex and the direction of the parabola (upward or downward).

STEP 2

1. Identify the coefficients $a$ , $b$ , and $c$ .
2. Find the vertex of the parabola.
3. Determine the axis of symmetry.
4. Determine the domain of the function.
5. Determine the range of the function.
6. Use the vertex and intercepts to sketch the graph.

STEP 3

Identify the coefficients $a$ , $b$ , and $c$ from the quadratic function $f(x) = 4x - x^2 + 12$ .
Rewriting it in standard form $f(x) = -x^2 + 4x + 12$ , we have: $a = -1, \quad b = 4, \quad c = 12$

STEP 4

Find the vertex of the parabola using the vertex formula $x = -\frac{b}{2a}$ .
$x = -\frac{4}{2(-1)} = 2$
Substitute $x = 2$ back into the function to find the y-coordinate of the vertex:
$f(2) = 4(2) - (2)^2 + 12 = 8 - 4 + 12 = 16$
Thus, the vertex is $(2, 16)$ .

STEP 5

Determine the axis of symmetry, which is the vertical line that passes through the vertex.
The axis of symmetry is $x = 2$ .

STEP 6

Determine the domain of the function.
Since quadratic functions are defined for all real numbers, the domain is:
$(-\infty, \infty)$

STEP 7

Determine the range of the function. Since the parabola opens downward (because $a = -1 < 0$ ), the range is from negative infinity to the y-coordinate of the vertex.
The range is:
$(-\infty, 16]$

STEP 8

To sketch the graph, use the vertex $(2, 16)$ and find the intercepts.
Find the y-intercept by setting $x = 0$ :
$f(0) = 4(0) - (0)^2 + 12 = 12$
The y-intercept is $(0, 12)$ .
Find the x-intercepts by setting $f(x) = 0$ :
$-x^2 + 4x + 12 = 0$
Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :
$x = \frac{-4 \pm \sqrt{4^2 - 4(-1)(12)}}{2(-1)}$ $x = \frac{-4 \pm \sqrt{16 + 48}}{-2}$ $x = \frac{-4 \pm \sqrt{64}}{-2}$ $x = \frac{-4 \pm 8}{-2}$
The solutions are $x = 6$ and $x = -2$ .
Thus, the x-intercepts are $(6, 0)$ and $(-2, 0)$ .
Use the vertex $(2, 16)$ and the intercepts to draw the graph.
The axis of symmetry is $x = 2$ .
The domain of the function is $(-\infty, \infty)$ .
The range of the function is $(-\infty, 16]$ .

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Identify the coefficients $a$ , $b$ , and $c$ .
2. Find the vertex of the parabola.
3. Determine the axis of symmetry.
4. Determine the domain of the function.
5. Determine the range of the function.
6. Use the vertex and intercepts to sketch the graph.

STEP 3

Identify the coefficients $a$ , $b$ , and $c$ from the quadratic function $f(x) = 4x - x^2 + 12$ .
Rewriting it in standard form $f(x) = -x^2 + 4x + 12$ , we have: $a = -1, \quad b = 4, \quad c = 12$

STEP 4

Find the vertex of the parabola using the vertex formula $x = -\frac{b}{2a}$ .
$x = -\frac{4}{2(-1)} = 2$
Substitute $x = 2$ back into the function to find the y-coordinate of the vertex:
$f(2) = 4(2) - (2)^2 + 12 = 8 - 4 + 12 = 16$
Thus, the vertex is $(2, 16)$ .

STEP 5

Determine the axis of symmetry, which is the vertical line that passes through the vertex.
The axis of symmetry is $x = 2$ .

STEP 6

Determine the domain of the function.
Since quadratic functions are defined for all real numbers, the domain is:
$(-\infty, \infty)$

STEP 7

Determine the range of the function. Since the parabola opens downward (because $a = -1 < 0$ ), the range is from negative infinity to the y-coordinate of the vertex.
The range is:
$(-\infty, 16]$

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Identify the coefficients a a a, b b b, and c c c.2. Find the vertex of the parabola.3. Determine the axis of symmetry.4. Determine the domain of the function.5. Determine the range of the function.6. Use the vertex and intercepts to sketch the graph.

STEP 3

Identify the coefficients a a a, b b b, and c c c from the quadratic function f(x)=4x−x2+12 f(x) = 4x - x^2 + 12 f(x)=4x−x2+12.Rewriting it in standard form f(x)=−x2+4x+12 f(x) = -x^2 + 4x + 12 f(x)=−x2+4x+12, we have: a=−1,b=4,c=12 a = -1, \quad b = 4, \quad c = 12 a=−1,b=4,c=12

STEP 4

STEP 5

Determine the axis of symmetry, which is the vertical line that passes through the vertex.The axis of symmetry is x=2 x = 2 x=2.

STEP 6

Determine the domain of the function.Since quadratic functions are defined for all real numbers, the domain is:(−∞,∞) (-\infty, \infty) (−∞,∞)

STEP 7

Determine the range of the function. Since the parabola opens downward (because a=−1<0 a = -1 < 0 a=−1<0), the range is from negative infinity to the y-coordinate of the vertex.The range is:(−∞,16] (-\infty, 16] (−∞,16]

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Identify the coefficients $a$ , $b$ , and $c$ .
2. Find the vertex of the parabola.
3. Determine the axis of symmetry.
4. Determine the domain of the function.
5. Determine the range of the function.
6. Use the vertex and intercepts to sketch the graph.

Identify the coefficients $a$ , $b$ , and $c$ from the quadratic function $f(x) = 4x - x^2 + 12$ .
Rewriting it in standard form $f(x) = -x^2 + 4x + 12$ , we have: $a = -1, \quad b = 4, \quad c = 12$

Determine the axis of symmetry, which is the vertical line that passes through the vertex.
The axis of symmetry is $x = 2$ .

Determine the domain of the function.
Since quadratic functions are defined for all real numbers, the domain is:
$(-\infty, \infty)$

Determine the range of the function. Since the parabola opens downward (because $a = -1 < 0$ ), the range is from negative infinity to the y-coordinate of the vertex.
The range is:
$(-\infty, 16]$