Math

QuestionGraph the function f(x)=(x1)2+7f(x)=(x-1)^{2}+7 using its vertex and yy-intercept to find the range.

Studdy Solution

STEP 1

Assumptions1. The function is given by f(x)=(x1)+7f(x)=(x-1)^{}+7 . The function is a quadratic function, which has a parabolic shape.
3. The vertex of the parabola is given by the point (h,k)(h, k) where hh is the value of xx that makes the square term equal to zero and kk is the value of the function at hh.
4. The y-intercept is the point where the graph of the function crosses the y-axis, i.e., when x=0x =0.

STEP 2

First, we find the vertex of the parabola. The vertex form of a quadratic function is f(x)=a(xh)2+kf(x) = a(x-h)^2 + k, where (h,k)(h, k) is the vertex of the parabola. In our case, h=1h =1 and k=7k =7.

STEP 3

So, the vertex of the parabola is (1,7)(1,7).

STEP 4

Next, we find the y-intercept of the graph. The y-intercept is the value of the function when x=0x =0.

STEP 5

Substitute x=0x =0 into the function to find the y-intercept.
f(0)=(01)2+7f(0) = (0-1)^2 +7

STEP 6

Calculate the y-intercept.
f(0)=1+=8f(0) =1 + =8

STEP 7

So, the y-intercept of the graph is (0,)(0,).

STEP 8

Now, we can draw the graph. Plot the vertex and the y-intercept on the graph.

STEP 9

Since the coefficient of x2x^2 is positive, the parabola opens upwards. Draw a smooth curve through the vertex and the y-intercept to complete the graph.

STEP 10

The range of the function is the set of all possible y-values. Since the parabola opens upwards and the vertex is at y=7y =7, the range of the function is [7,+)[7, +\infty).
The graph of the function f(x)=(x)2+7f(x)=(x-)^{2}+7, the vertex of the parabola is (,7)(,7), the y-intercept is (0,8)(0,8), and the range is [7,+)[7, +\infty).

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