Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
2. Quadratic Functions
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Exercise 24 Page 557

Start by identifying the values of a, b, and c.

Graph:

Axis of Symmetry: x=- 2.5
Vertex: (- 2.5,12.5)

Practice makes perfect

To draw the graph of the given quadratic function written in standard form, we must start by identifying the values of a, b, and c. y=- 2x^2-10x ⇔ y=( - 2)x^2+( - 10)x+ 0 We can see that a= - 2, b= - 10, and c= 0. Now, we will follow four steps to graph the function.

  1. Find the axis of symmetry.
  2. Calculate the vertex.
  3. Identify the y-intercept and its reflection across the axis of symmetry.
  4. Connect the points with a parabola.

    Finding the Axis of Symmetry

    The axis of symmetry is a vertical line with equation x= - b2 a. Since we already know the values of a and b, we can substitute them into the formula.
    x= - b/2a
    x= - ( - 10)/2( - 2)
    â–Ľ
    Simplify right-hand side
    x=-(- 10)/- 2 * 2
    x=10/- 4
    x=- 2.5
    The axis of symmetry of the parabola is the vertical line with equation x=- 2.5.

    Calculating the Vertex

    To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: ( - b/2 a, f( - b/2 a ) ) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=- 2.5. Therefore, the x-coordinate of the vertex is also - 2.5. To find the y-coordinate, we need to substitute - 2.5 for x in the given equation.
    y = - 2x^2-10x
    y = - 2( - 2.5)^2-10( - 2.5)
    â–Ľ
    Simplify right-hand side
    y = - 2(6.25)-10(- 2.5)
    y = - 2(6.25)-(- 25)
    y = - 12.5 -(- 25)
    y = - 12.5 + 25
    y = 12.5
    We found the y-coordinate, and now we know that the vertex is (- 2.5,12.5).

    Identifying the y-intercept and Its Reflection

    The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0, 0). Let's plot this point and its reflection across the axis of symmetry.

    Connecting the Points

    We can now draw the graph of the function. Since a= - 2, which is negative, the parabola will open downwards. Let's connect the three points with a smooth curve.