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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 11 Page 534

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

x=- 4

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.

- 8x-16=x^2

LHS-x^2=RHS-x^2

- 8x-16-x^2=0

CommutativePropAdd

Commutative Property of Addition

- x^2-8x-16=0

Now we can identify the function related to the equation. Equation:&- x^2-8x-16=0 Related Function:&y=- x^2-8x-16

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^2-8x-16 ⇔ y= - 1x^2+( - 8)x+( - 16) We can see that a= - 1, b= - 8, and c= - 16. 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=- b2 a. Since we already know the values of a and b, we can substitute them into the formula.

x=- b/2a

a= - 1, b= - 8

x=- - 8/2( - 1)

▼

Simplify right-hand side

a(- b)=- a * b

x=-- 8/- 2

- a/- b=a/b

x=-8/2

Calculate quotient

x=- 4

The axis of symmetry of the parabola is the vertical line with equation x=- 4.

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=- 4. Thus, the x-coordinate of the vertex is also - 4. To find the y-coordinate, we need to substitute - 4 for x in the given equation.

y=- x^2-8x-16

x= - 4

y=- ( - 4)^2-8( - 4)-16

▼

Simplify right-hand side

NegBaseToPosBase

(- a)^2=a^2

y=- 16-8(- 4)-16

MultNegNegOnePar

- a(- b)=a* b

y=- 16+32-16

Add and subtract terms

y=0

We found the y-coordinate, and now we know that the vertex is (- 4,0).

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, - 16). 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 negative, the parabola will open downwards. 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 intercepts the x-axis once. The point of intersection is ( - 4,0), it is also the vertex of the parabola. The equation - 8x-16=x^2 has one solution, x= - 4.

Checking Our Answer

Checking Our Solutions

We can check our solution by substituting the value into the given equation.

- 8x-16=x^2

x= - 4

- 8( - 4)-16? =( - 4)^2

▼

Simplify

MultNegNegOnePar

- a(- b)=a* b

32-16? =(- 4)^2

Subtract term

16? =(- 4)^2

NegBaseToPosBase

(- a)^2=a^2

16=16 ✓

Since 16=16, we know that x=- 4 is a solution of the equation.

Subchapter links

Exercises p.534-536