Chapter 2.1, Problem 16E

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 + 5

To determine

To graph: The function f(x)=x2+5 and indicate the coordinates of the vertex, the y-intercept and x-intercept.

Explanation

Given information:

The provided function is:

f(x)=āx2+5

Graph:

Consider the function,

f(x)=āx2+5

Compare the equation f(x)=āx2+5 with the standard function f(x)=ax2+bx+c and find the value of a,b and c.

The values are a=ā1,b=0 and c=5.

Since, a<0 therefore the parabola will be downward 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=ā1 and b=0 in the equation x=āb2a.

x=ā(0)2(ā1)=0

The y-coordinate of vertex,

Substitute x=0 in the function f(x)=āx2+5.

f(x)=ā(0)2+5=5

Thus, coordinates of the vertex are (0,5).

x-intercept: To calculate x-intercept of the function, substitute f(x)=0 in the equation f(x)=āx2+5 and solve.

āx2+5=0

To find the value of x use the formula,

x=ābĀ±b2ā4ac2a

Substitute a=ā1,b=0 and c=5 in the equation x=ābĀ±b2ā4ac2a.

x=ā(0)Ā±(0)2ā4(1)(5)2(ā1)=Ā±ā202

For real values, square root of a negative number does not exist therefore there will not be any x-intercept

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