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Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

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Chapter
Section
BuyFindarrow_forward

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)=x2+5

Graph:

Consider the function,

f(x)=x2+5

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

The values are a=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=b2a

Substitute a=1 and b=0 in the equation x=b2a.

x=(0)2(1)=0

The y-coordinate of vertex,

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

f(x)=(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)=x2+5 and solve.

x2+5=0

To find the value of x use the formula,

x=b±b24ac2a

Substitute a=1,b=0 and c=5 in the equation x=b±b24ac2a.

x=(0)±(0)24(1)(5)2(1)=±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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