Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
Chapter Review
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Exercise 8 Page 84

Start by identifying the axis of symmetry. Then use it to find the vertex and the maximum value of g.

Minimum Value: - 25
Decreasing Interval: To the left of x=- 2
Increasing Interval: To the right of x=- 2

Practice makes perfect

Note that the function is written in intercept form, h(x)=a(x-p)(x-q) where a, p, and q are either positive or negative numbers. h(x)=(x-3)(x+7) To solve the problem and draw the graph, we will follow four steps.

  1. Identify the constants a, p, and q.
  2. Plot the vertex and draw the axis of symmetry.
  3. Find the minimum or maximum value of h and describe where the function is increasing and decreasing.
  4. Sketch the curve.

Let's get started.

Step 1

We will first identify the constants a, p, and q. Recall that if a<0, the parabola will open downwards. Conversely, if a>0, the parabola will open upwards. Intercept form:& h(x)=a(x-p)(x-q) Function:& h(x)=1(x-3)(x-(- 7)) We can see that a=1, p=3, and q=- 7. Since a is grater than 0, the parabola will open upwards.

Step 2

The axis of symmetry is a vertical line with equation x= p+q2. Since we already know the values of p and q, we can substitute them into the formula.

x=p+q/2
x=3+(- 7)/2
â–¼
Simplify right-hand side
x=- 4/2
x=- 2

The axis of symmetry of the parabola is the vertical line with equation x=- 2. To calculate the vertex, we can write the expression for the vertex by stating the x- and y-coordinates in terms of p and q. Vertex:& (p+q/2, h (p+q/2 ) ) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=- 2. Thus, the x-coordinate of the vertex is also - 2. To find the y-coordinate, we need to substitute - 2 for x in the given equation.

h(x)=(x-3)(x+7)
h(- 2)=(- 2-3)(- 2+7)
â–¼
Simplify right-hand side
h(2)=- 5 * 5
h(2)=- 25

We found the y-coordinate, and now we know that the vertex is (- 2,- 25).

Step 3

Since a=1 is greater than 0, the function decreases to the left of the minimum value and increases to the right of the minimum value, which we know occurs at x=- 2. This means that h(- 2)=- 25 will be the maximum value of h. Minimum value:& - 25 Decreasing Interval:& To the left of - 2 Increasing Interval:& To the right of - 2

Step 4

We will now plot a point on the curve by choosing an x-value and calculating its corresponding y-value. Let's try x=2.

h(x)=(x-3)(x+7)
h(2)=(2-3)(2+7)
â–¼
Simplify right-hand side
h(2)=(- 1) 9
h(2)=- 9

When x=2, we have h(2)=- 9. Thus, the point (2,- 9) lies on the curve. Let's plot this point and reflect it across the axis of symmetry.

Note that both points have the same y-coordinate. Finally, we will sketch the parabola which passes through the three points. Remember not to use a straightedge for this!