Exercise 146 Page 537

a We want to identify the vertex and decide whether it is the maximum or minimum point of the given quadratic function. Note that the formula is already expressed in graphing form, f(x)=a(x-h)^2+k, where a, h, and k are either positive or negative numbers.

f(x)=(x-2)^2+9

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- 2)^2+9 We can see that a= 1, h= 2, and k=9. The vertex of a quadratic function written in graphing form is the point ( h,k). For this exercise, we have h= 2 and k=9. Therefore, the vertex of the given equation is ( 2,9).

Maximum or Minimum Value

Before we determine if the vertex is the maximum or minimum point recall that, if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards.

In the given function, we have a= 1, which is greater than 0. Thus, the parabola opens upwards and we will have a minimum value. The vertex is always the lowest or the highest point on the graph. Therefore, in this case, the vertex represents the minimum value of the function.

b Let's analyze the given function. It contains two parts: a square and a constant ( c).

f(x)= (x-2)^2_(square)+ 9 ← constant Notice that the constant is positive and regardless of what number we substitute for x, the square part will always be non-negative. Adding a positive constant to non-negative number will always give us a positive value of a function, so the graph of the given function will be above the x-axis. Let's look at the given equation. (x-2)^2+9=0 This is the equation for the x-intercepts of the given function. Since the graph is above the x-axis, there are no x-intercepts. Therefore, the equation has no real solutions.

c To find the roots of a function — real or imaginary — we need to set the function equal to 0 and then solve for x.

(x-2)^2+9=0We want to solve the above quadratic equation to find the imaginary roots. Let's begin by isolating the square on one side of the equation. (x-2)^2+9=0 ⇔ (x-2)^2=- 9 To solve the above equation we will take the square root of both sides of the equation. Since this method gives two solutions — a negative and a positive — remember to consider them both by adding ± to the solution.

(x-2)^2=- 9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((x-2)^2)=sqrt(- 9)

CalcRoot

Calculate root

x-2=± sqrt(- 9)

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Solve for x

AddEqn

LHS+2=RHS+2

x=2 ± sqrt(- 9)