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QuestionSketch the graph of the quadratic function f(x)=(x+2)2−9f(x)=(x+2)^{2}-9. Find the axis of symmetry, domain, and range.

Studdy Solution

STEP 1

Assumptions1. The given function is a quadratic function in the form f(x)=a(x−h)+kf(x)=a(x-h)^{}+k, where (h, k) is the vertex of the parabola. . The function is f(x)=(x+)−9f(x)=(x+)^{}-9.
3. The vertex form of a parabola is used to easily identify the vertex and axis of symmetry of the parabola.
4. The domain of a function is the set of all possible x-values and the range is the set of all possible y-values.
5. The axis of symmetry of a parabola is the vertical line through the vertex.
6. The intercepts are the points where the graph intersects the x-axis and y-axis.

STEP 2

First, we need to find the vertex of the parabola. The vertex is given by the point (h, k) in the equation f(x)=a(x−h)2+kf(x)=a(x-h)^{2}+k.
Vertex=(−h,k)Vertex = (-h, k)

STEP 3

Now, plug in the given values for h and k to find the vertex.
Vertex=(−(−2),−9)Vertex = (-(-2), -9)

STEP 4

Calculate the vertex.
Vertex=(2,−9)Vertex = (2, -9)

STEP 5

Next, we need to find the axis of symmetry. The axis of symmetry is the line x=hx = h.
Axis of symmetry=x=hAxis\, of\, symmetry = x = h

STEP 6

Plug in the value for h to find the axis of symmetry.
Axis of symmetry=x=−(−2)Axis\, of\, symmetry = x = -(-2)

STEP 7

Calculate the axis of symmetry.
Axis of symmetry=x=2Axis\, of\, symmetry = x =2

STEP 8

Now, we need to find the intercepts. The x-intercepts are found by setting f(x)=0f(x) =0 and solving for x.
0=(x+2)2−0 = (x+2)^{2}-

STEP 9

olve the equation for x to find the x-intercepts.
x=−2±9x = -2 \pm \sqrt{9}

STEP 10

Calculate the x-intercepts.
x=−2±3x = -2 \pm3x=,−5x =, -5

STEP 11

The y-intercept is found by setting x=0x =0 and solving for y.
y=(0+)−9y = (0+)^{}-9

STEP 12

Calculate the y-intercept.
y=4−9=−5y =4 -9 = -5

STEP 13

Now that we have the vertex, axis of symmetry, and intercepts, we can sketch the graph of the function.

STEP 14

The domain of a quadratic function is all real numbers, so the domain of this function is (−∞,∞)(-\infty, \infty).

STEP 15

The range of a quadratic function is all y-values greater than or equal to the y-coordinate of the vertex for a parabola that opens upwards, and all y-values less than or equal to the y-coordinate of the vertex for a parabola that opens downwards. Since our parabola opens upwards, the range is (−∞,k](-\infty, k].

STEP 16

Plug in the value for k to find the range.
Range=(−∞,−9]Range = (-\infty, -9]The vertex of the parabola is (2, -9), the axis of symmetry is x=2x =2, the x-intercepts are and -5, the y-intercept is -5, the domain is (−∞,∞)(-\infty, \infty), and the range is (−∞,−9](-\infty, -9].

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