Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
2. Solving Quadratic Equations by Graphing
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Exercise 1 Page 490

Make sure the equation is written in standard form. Identify the related function and graph it.

x=-1, x=2

Practice makes perfect

We are asked to solve the given quadratic equation by graphing. There are three steps to solving a quadratic equation by graphing.

  1. Write the equation in standard form, ax^2+bx+c=0.
  2. Graph the related function y=ax^2+bx+c.
  3. Find the x-intercepts, if any.

The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Our equation is already written in standard form. Let's identify the function related to the equation. Equation:&x^2-x-2=0 Related Function:&y=x^2-x-2

Graphing the Related Function

To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c.

y=x^2-x-2 ⇔ y=1x^2+(- 1)x+(-2) We can see that a=1, b=- 1, and c=-2. 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 the equation x=- b2a. Since we already know the values of a and b, we can substitute them into the formula.
x=- b/2a
x=- - 1/2(1)
â–Ľ
Simplify right-hand side
x=-- 1/2
x=1/2
x=0.5
The axis of symmetry for the parabola is a vertical line with the equation x=0.5.

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/2a, f( - b/2a ) ) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=0.5. Therefore, the x-coordinate of the vertex is also 0.5. To find the y-coordinate, we need to substitute 0.5 for x in the given equation.
y=x^2-x-2
y= 0.5^2- 0.5-2
â–Ľ
Simplify right-hand side
y=0.25-0.5-2
y=- 2.25
We found the y-coordinate and now we know that the vertex is (0.5,- 2.25).

Identifying the y-intercept and its Reflection

The y-intercept of the graph for a quadratic function written in standard form is given by the value of c. The point where our graph intercepts the y-axis is (0,-2). 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=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.

Finding the x-intercepts

Let's identify the x-intercepts of the graph of the related function.

We can see that the parabola intersects the x-axis twice. The points of intersection are ( -1,0) and ( 2,0). Therefore, the equation x^2-x-2=0 has two solutions, x= -1 and x= 2.

Checking Our Answer

Checking Our Solutions
We can check our solutions by substituting the values into the given equation. If our solutions are correct, the final result will be 0=0. Let's first check x=-1.
x^2-x-2=0
( -1)^2-( -1)-2? =0
â–Ľ
Simplify left-hand side
1-(-1)-2? =0
1+1-2? =0
0=0 âś“
Great! Now, let's check our second value, x=2.
x^2-x-2=0
2^2- 2-2? =0
â–Ľ
Simplify left-hand side
4-2-2? =0
0=0 âś“