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Big Ideas Math Algebra 1, 2015

BI

Big Ideas Math Algebra 1, 2015 View details

Chapter Review

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Exercise 10 Page 534

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

No solutions

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.

Write the equation in standard form, ax^2+bx+c=0.
Graph the related function y=ax^2+bx+c.
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. Let's write our equation in standard form. This means gathering all of the terms on the left-hand side of the equation. x^2-2x=- 4 ⇔ x^2-2x+4=0 Now we can identify the function related to the equation. Equation:&x^2-2x+4=0 Related Function:&y=x^2-2x+4

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^2-2x+4 ⇔ y=1x^2+(- 2)x+4 We can see that a=1, b=- 2, and c=4. Now, we will follow four steps to graph the function.

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

Finding the Axis of Symmetry

The axis of symmetry is a vertical line with equation x=- b2a. Since we already know the values of a and b, we can substitute them into the formula.

x=- b/2a

a= 1, b= - 2

x=- - 2/2(1)

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Simplify right-hand side

a * 1=a

x=-- 2/2

RemoveNegFracAndNum

- - a/b= a/b

x=2/2

a/a=1

x=1

The axis of symmetry of the parabola is the vertical line with equation x=1.

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=1. Thus, the x-coordinate of the vertex is also 1. To find the y-coordinate, we need to substitute 1 for x in the given equation.

y=x^2-2x+4

x= 1

y= 1^2-2( 1)+4

▼

Simplify right-hand side

1^a=1

y=1-2(1)+4

a * 1=a

y=1-2+4

Add and subtract terms

y=3

We found the y-coordinate, and now we know that the vertex is (1,3).

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 (0,4). 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, if there are any.

We can see that the parabola does not intercept the x-axis. Therefore, the equation x^2-2x=- 4 has no solutions.

Subchapter links

Exercises p.534-536