Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
2. Characteristics of Quadratic Functions
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Exercise 59 Page 63

Identify the vertex first. Then use it to find the axis of symmetry.

x-intercept: x=3
Axis of Symmetry: x=3
Vertex: (3,0)
Graph:

Practice makes perfect

We want to draw the graph of the given quadratic function. To do so, we will rewrite it in vertex form, f(x)=a(x-h)^2+k, where a, h, and k are either positive or negative numbers. f(x)=-2(x-3)^2 ⇕ f(x)=-2(x-3)^2+0 To draw the graph, we will follow four steps.

  1. Identify the constants a, h, and k.
  2. Plot the vertex (h,k) and draw the axis of symmetry x=h.
  3. Plot any point on the curve and its reflection across the axis of symmetry.
  4. Sketch the curve.

    Let's get started.

    Step 1

    We will first identify the constants a, h, and k. Recall that if a<0, the parabola will open downwards. Conversely, if a>0, the parabola will open upwards. Vertex Form:& f(x)= a(x- h)^2+k Function:& f(x)= -2(x- 3)^2+ We can see that a= -2, h= 3, and k= . Since a is less than 0, the parabola will open downwards.

    Step 2

    Let's now plot the vertex ( h,k) and draw the axis of symmetry x= h. Since we already know the values of h and k, we know that the vertex is ( 3, ). Therefore, the axis of symmetry is the vertical line x= 3.

    Step 3

    We will now plot a point on the curve by choosing an x-value and calculating its corresponding y-value. Let's try x=1.
    f(x)=-2(x-3)^2
    f( 1)=-2( 1-3)^2
    â–Ľ
    Simplify right-hand side
    f(1)=-2(-2)^2
    f(1)=-2(4)
    f(1)=-8
    When x=1, we have y=- 8. Thus, the point (1,- 8) 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.

    Step 4

    Finally, we will sketch the parabola which passes through the three points. Remember not to use a straightedge for this!