Exercise 123 Page 593

a We want to identify the vertex of the given quadratic function. To do so we will first express it in graphing form, f(x)=a(x-h)^2+k, where a, h, and k are either positive or negative constants.

f(x)=(x-1)^2-4 ⇕ f(x)=1(x-1)^2+(-4) It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the graphing form to our equation. General Formula:f(x)=& a(x- h)^2 +k Equation:f(x)=& 1(x- 1)^2+(-4) We can see that a= 1, h= 1, and k=- 4. The vertex of a quadratic function written in graphing form is the point ( h,k). For this exercise, we have h= 1 and k=-4. Therefore, the vertex of the given equation is ( 1,- 4).

b We want to determine the x-intercepts of the given function. Let's start by rewriting the function by substituting y in place of f(x).

f(x)=(x-1)^2-4 ⇕ y=(x-1)^2-4 Think of the point where the graph of an equation crosses the x-axis. The y-value of that ( x, y) coordinate pair is 0, and the x-value is the x-intercept. To find the x-intercepts of the function, we should substitute 0 for y and solve for x.

y=(x-1)^2-4

Substitute

y= 0

0=(x-1)^2-4

▼

Solve for x

AddEqn

LHS+4=RHS+4

4=(x-1)^2

RearrangeEqn

Rearrange equation

(x-1)^2=4

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-1)^2)=sqrt(4)

CalcRoot

Calculate root

x-1=± 2

AddEqn

LHS+1=RHS+1

x=1 ± 2

Both x=1+2= 3 and x=1-2= - 1 are the x-intercepts of the given function. It means that the graph passes through the x-axis at the points ( 3,0) and ( - 1,0).

c We want to sketch a graph of the given function. From Part B we already know that the x-intercepts occur at (3,0) and (- 1,0). In Part A we also found that the given function has the vertex in the point (1,- 4). Let's plot these points!

We can now draw the parabola through the vertex and the x-intercepts.

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Exercise 123 Page 593

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