Chapter 2.1, Problem 10E

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Chapter
Section

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

# In Exercises 7–16, sketch the graph of the quadratic function, indicating the coordinates of the vertex, the y-intercept, and the x-intercepts (if any). [HINT: See Example 1.] f ( x ) = x 2 + 2 x + 1

To determine

To graph: The function f(x)=x2+2x+1 and indicate the coordinates of the vertex, the y-intercept and x-intercept.

Explanation

Given information:

The provided function is:

f(x)=x2+2x+1

Graph:

Consider the function,

f(x)=x2+2x+1

Compare the equation f(x)=x2+2x+1 with the standard function f(x)=ax2+bx+c and find the value of a,b and c.

The values are a=1,b=2 and c=1.

Since, a>0 therefore the parabola will be upward facing,

To graph a quadratic function, four things should be calculated first.

(i) Vertex

(ii) x-intercept

(iii) y- intercept

(iv) symmetry

Vertex: The formula of x- coordinate of vertex is,

x=āb2a

Substitute a=1 and b=2 in the equation x=āb2a.

x=ā(2)2(1)=ā1

The y-coordinate of vertex,

Substitute x=ā1 in the function f(x)=x2+2x+1.

f(x)=(ā1)2+2(ā1)+1=1ā2+1=0

Thus, coordinates of the vertex are (ā1,0).

x-intercept: To calculate x-intercept of the function, substitute f(x)=0 in the equation f(x)=x2+2x+1 and solve.

x2+2x+1=0x2+x+x+1=0x(x+1)+1(x+1)=0(x+1)(x+1)=0

Now solve for x+1

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