Math / Algebra

Question9) $f(x)=(x+2)(x-2)$
Minimum / Maximum Domain: $\qquad$ Range: $\qquad$ x-intercepts: $\qquad$ y-intercept: $\qquad$ Vertex: $\qquad$

Studdy Solution

STEP 1

1. The function $f(x) = (x+2)(x-2)$ is a quadratic function.
2. The quadratic function can be expressed in standard form $ax^2 + bx + c$ .
3. The domain of any polynomial function is all real numbers.
4. The vertex form of a quadratic function helps in finding the vertex and determining if it is a minimum or maximum.
5. The x-intercepts occur where $f(x) = 0$ .
6. The y-intercept occurs where $x = 0$ .

STEP 2

1. Express the function in standard form.
2. Determine the domain of the function.
3. Determine the range of the function.
4. Find the x-intercepts.
5. Find the y-intercept.
6. Find the vertex and determine if it is a minimum or maximum.

STEP 3

Expand the function to express it in standard form:
$f(x) = (x+2)(x-2) = x^2 - 4$

STEP 4

Determine the domain of the function. Since $f(x)$ is a polynomial, its domain is all real numbers:
$\text{Domain: } (-\infty, \infty)$

STEP 5

Determine the range of the function. Since the parabola opens upwards (the coefficient of $x^2$ is positive), the minimum value is at the vertex, and the range is from the vertex value to infinity:
$\text{Range: } [-4, \infty)$

STEP 6

Find the x-intercepts by setting $f(x) = 0$ :
$(x+2)(x-2) = 0$
The solutions are:
$x = -2 \quad \text{and} \quad x = 2$
$\text{x-intercepts: } x = -2, x = 2$

STEP 7

Find the y-intercept by evaluating $f(x)$ at $x = 0$ :
$f(0) = (0+2)(0-2) = -4$
$\text{y-intercept: } y = -4$

STEP 8

Find the vertex of the parabola. The vertex form of a quadratic is $f(x) = a(x-h)^2 + k$ , where $(h, k)$ is the vertex. For $f(x) = x^2 - 4$ , the vertex is at the point where $x = 0$ :
$\text{Vertex: } (0, -4)$
Since the parabola opens upwards, the vertex is a minimum.
$\text{Minimum: } (0, -4)$
The solution is: - Domain: $(-\infty, \infty)$ - Range: $[-4, \infty)$ - x-intercepts: $x = -2, x = 2$ - y-intercept: $y = -4$ - Vertex: $(0, -4)$ - Minimum: $(0, -4)$

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Question9) $f(x)=(x+2)(x-2)$
Minimum / Maximum Domain: $\qquad$ Range: $\qquad$ x-intercepts: $\qquad$ y-intercept: $\qquad$ Vertex: $\qquad$

Studdy Solution

STEP 1

STEP 2

1. Express the function in standard form.
2. Determine the domain of the function.
3. Determine the range of the function.
4. Find the x-intercepts.
5. Find the y-intercept.
6. Find the vertex and determine if it is a minimum or maximum.

STEP 3

Expand the function to express it in standard form:
$f(x) = (x+2)(x-2) = x^2 - 4$

STEP 4

Determine the domain of the function. Since $f(x)$ is a polynomial, its domain is all real numbers:
$\text{Domain: } (-\infty, \infty)$

STEP 5

Determine the range of the function. Since the parabola opens upwards (the coefficient of $x^2$ is positive), the minimum value is at the vertex, and the range is from the vertex value to infinity:
$\text{Range: } [-4, \infty)$

STEP 6

Find the x-intercepts by setting $f(x) = 0$ :
$(x+2)(x-2) = 0$
The solutions are:
$x = -2 \quad \text{and} \quad x = 2$
$\text{x-intercepts: } x = -2, x = 2$

STEP 7

Find the y-intercept by evaluating $f(x)$ at $x = 0$ :
$f(0) = (0+2)(0-2) = -4$
$\text{y-intercept: } y = -4$

STEP 8

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Question9) f(x)=(x+2)(x−2)f(x)=(x+2)(x-2)f(x)=(x+2)(x−2)Minimum / Maximum Domain: \qquad Range: \qquad x-intercepts: \qquad y-intercept: \qquad Vertex: \qquad

Studdy Solution

STEP 1

STEP 2

1. Express the function in standard form.2. Determine the domain of the function.3. Determine the range of the function.4. Find the x-intercepts.5. Find the y-intercept.6. Find the vertex and determine if it is a minimum or maximum.

STEP 3

Expand the function to express it in standard form:f(x)=(x+2)(x−2)=x2−4 f(x) = (x+2)(x-2) = x^2 - 4 f(x)=(x+2)(x−2)=x2−4

STEP 4

Determine the domain of the function. Since f(x) f(x) f(x) is a polynomial, its domain is all real numbers:Domain: (−∞,∞) \text{Domain: } (-\infty, \infty) Domain: (−∞,∞)

STEP 5

Determine the range of the function. Since the parabola opens upwards (the coefficient of x2 x^2 x2 is positive), the minimum value is at the vertex, and the range is from the vertex value to infinity:Range: [−4,∞) \text{Range: } [-4, \infty) Range: [−4,∞)

STEP 6

Find the x-intercepts by setting f(x)=0 f(x) = 0 f(x)=0:(x+2)(x−2)=0 (x+2)(x-2) = 0 (x+2)(x−2)=0The solutions are:x=−2andx=2 x = -2 \quad \text{and} \quad x = 2 x=−2andx=2x-intercepts: x=−2,x=2 \text{x-intercepts: } x = -2, x = 2 x-intercepts: x=−2,x=2

STEP 7

Find the y-intercept by evaluating f(x) f(x) f(x) at x=0 x = 0 x=0:f(0)=(0+2)(0−2)=−4 f(0) = (0+2)(0-2) = -4 f(0)=(0+2)(0−2)=−4y-intercept: y=−4 \text{y-intercept: } y = -4 y-intercept: y=−4

STEP 8

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Question9) $f(x)=(x+2)(x-2)$
Minimum / Maximum Domain: $\qquad$ Range: $\qquad$ x-intercepts: $\qquad$ y-intercept: $\qquad$ Vertex: $\qquad$

1. Express the function in standard form.
2. Determine the domain of the function.
3. Determine the range of the function.
4. Find the x-intercepts.
5. Find the y-intercept.
6. Find the vertex and determine if it is a minimum or maximum.

Expand the function to express it in standard form:
$f(x) = (x+2)(x-2) = x^2 - 4$

Determine the domain of the function. Since $f(x)$ is a polynomial, its domain is all real numbers:
$\text{Domain: } (-\infty, \infty)$

Determine the range of the function. Since the parabola opens upwards (the coefficient of $x^2$ is positive), the minimum value is at the vertex, and the range is from the vertex value to infinity:
$\text{Range: } [-4, \infty)$

Find the x-intercepts by setting $f(x) = 0$ :
$(x+2)(x-2) = 0$
The solutions are:
$x = -2 \quad \text{and} \quad x = 2$
$\text{x-intercepts: } x = -2, x = 2$

Find the y-intercept by evaluating $f(x)$ at $x = 0$ :
$f(0) = (0+2)(0-2) = -4$
$\text{y-intercept: } y = -4$