Graph f(x)=(x+4)^2

Problem

ƒ(x)=(x+4)2

Solution

1. Identify the parent function and the type of graph. The function ƒ(x)=(x+4)2 is a quadratic function, which means its graph is a parabola.

2. Determine the vertex by looking at the vertex form a*(x−h)2+k Here, h=−4 and k=0 so the vertex is at (−4,0)

3. Find the y-intercept by evaluating the function at x=0

ƒ(0)=(0+4)2

ƒ(0)=16

The y-intercept is (0,16)

4. Find the x-intercept by setting ƒ(x)=0

(x+4)2=0

x+4=0

x=−4

The x-intercept is the vertex, (−4,0)

5. Identify the axis of symmetry, which is the vertical line passing through the vertex.

x=−4

6. Plot additional points to define the shape. For x=−3

ƒ*(−3)=(−3+4)2=1=1

For x=−5

ƒ*(−5)=(−5+4)2=(−1)2=1

The points are (−3,1) and (−5,1)

7. Sketch the parabola opening upwards through the points (−5,1) (−4,0) and (−3,1) extending toward the y-intercept at (0,16)

Final Answer

ƒ(x)=(x+4)2* is a parabola with vertex *(−4,0)* opening upwards.

Want more problems? Check here!