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Find the intercepts of the parabola y = 9 x 2 + 12 x + 4 .

y : ( 0 , 4 ) ; x : ( 2 3 , 0 )

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Graph quadratic equations in two variables

Now, we have all the pieces we need in order to graph a quadratic equation in two variables. We just need to put them together. In the next example, we will see how to do this.

How to graph a quadratic equation in two variables

Graph y = x 2 6 x + 8 .

Solution

The image shows the steps to graph the quadratic equation y equals x squared minus 6 x plus 8. Step 1 is to write the quadratic equation with y on one side. This equation has y on one side already. The value of a is one, the value of b is -6 and the value of c is 8. Step 2 is to determine whether the parabola opens upward or downward. Since a is positive, the parabola opens upward. Step 3 is to find the axis of symmetry. The axis of symmetry is the line x equals negative b divided by the quantity 2 a. Plugging in the values of b and a the formula becomes x equals negative -6 divided by the quantity 2 times 1 which simplifies to x equals 3. The axis of symmetry is the line x equals 3. Step 4 is to find the vertex. The vertex is on the axis of symmetry. Substitute x equals 3 into the equation and solve for y. The equation is y equals x squared minus 6 x plus 8. Replacing x with 3 it becomes y equals 3 squared minus 6 times 3 plus 8 which simplifies to y equals -1. The vertex is (3, -1). Step 5 is to find the y-intercept and find the point symmetric to the y-intercept across the axis of symmetry. We substitute x equals 0 into the equation. The equation is y equals x squared minus 6 x plus 8. Replacing x with 0 it becomes y equals 0 squared minus 6 times 0 plus 8 which simplifies to y equals 8. The y-intercept is (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. The point symmetric to the y-intercept is (6, 8). Step 6 is to find the x-intercepts. We substitute y equals 0 into the equation. The equation becomes 0 equals x squared minus 6 x plus 8. We can solve this quadratic equation by factoring to get 0 equals the quantity x minus 2 times the quantity x minus 4. Solve each equation to get x equals 2 and x equals 4. The x-intercepts are (2, 0) and (4, 0). Step 7 is to graph the parabola. We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these five points to sketch the parabola. The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -2 to 10. The y-axis of the plane runs from -3 to 10. The vertex is at the point (3, -1). Four points are plotted on the curve at (0, 8), (6, 8), (2, 0) and (4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3.
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Graph the parabola y = x 2 + 2 x 8 .

y : ( 0 , −8 ) ; x : ( 2 , 0 ) , ( −4 , 0 ) ;
axis: x = −1 ; vertex: ( −1 , −9 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (-1, -9). Three points are plotted on the curve at (0, -8), (2, 0) and (-4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1.

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Graph the parabola y = x 2 8 x + 12 .

y : ( 0 , 12 ) ; x : ( 2 , 0 ) , ( 6 , 0 ) ;
axis: x = 4 ; vertex: ( 4 , −4 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (4, -4). Three points are plotted on the curve at (0, 12), (2, 0) and (6, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 4.

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Graph a quadratic equation in two variables.

  1. Write the quadratic equation with y on one side.
  2. Determine whether the parabola opens upward or downward.
  3. Find the axis of symmetry.
  4. Find the vertex.
  5. Find the y -intercept. Find the point symmetric to the y -intercept across the axis of symmetry.
  6. Find the x -intercepts.
  7. Graph the parabola.

We were able to find the x -intercepts in the last example by factoring. We find the x -intercepts in the next example by factoring, too.

Graph y = x 2 + 6 x 9 .

Solution

The equation y has on one side. .
Since a is 1 , the parabola opens downward.

To find the axis of symmetry, find x = b 2 a .
. .
.
.
The axis of symmetry is x = 3 . The vertex is on the line x = 3 .
.
Find y when x = 3 . .
.
.
.
The vertex is ( 3 , 0 ) .
.
The y -intercept occurs when x = 0 .
Substitute x = 0 .
Simplify.

The point ( 0 , −9 ) is three units to the left of the line of symmetry.
The point three units to the right of the line of symmetry is ( 6 , −9 ) .
Point symmetric to the y- intercept is ( 6 , −9 )
.
.
.
The y -intercept is ( 0 , −9 ) .
.
The x -intercept occurs when y = 0 . .
Substitute y = 0 . .
Factor the GCF. .
Factor the trinomial. .
Solve for x . .
Connect the points to graph the parabola. .
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Graph the parabola y = −3 x 2 + 12 x 12 .

y : ( 0 , −12 ) ; x : ( 2 , 0 ) ;
axis: x = 2 ; vertex: ( 2 , 0 ) ;
The graph shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -1 to 10. The vertex is at the point (2, 0). One other point is plotted on the curve at (0, -12). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2.

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Graph the parabola y = 25 x 2 + 10 x + 1 .

y : ( 0 , 1 ) ; x : ( 1 5 , 0 ) ;
axis: x = 1 5 ; vertex: ( 1 5 , 0 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (-1 fifth, 0). One other point is plotted on the curve at (0, 1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1 fifth.

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For the graph of y = x 2 + 6 x 9 , the vertex and the x -intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation 0 = x 2 + 6 x 9 is 0, so there is only one solution. That means there is only one x -intercept, and it is the vertex of the parabola.

How many x -intercepts would you expect to see on the graph of y = x 2 + 4 x + 5 ?

Graph y = x 2 + 4 x + 5 .

Solution

The equation has y on one side. .
Since a is 1, the parabola opens upward. .
To find the axis of symmetry, find x = b 2 a . .
.
.
The axis of symmetry is x = −2 .
.
The vertex is on the line x = −2 .
Find y when x = −2 . .
.
.
.
The vertex is ( −2 , 1 ) .
.
The y -intercept occurs when x = 0 .
Substitute x = 0 .
Simplify.
The point ( 0 , 5 ) is two units to the right of the line of symmetry.
The point two units to the left of the line of symmetry is ( −4 , 5 ) .
.
.
.
The y -intercept is ( 0 , 5 ) .
.
Point symmetric to the y- intercept is ( −4 , 5 ) .
The x - intercept occurs when y = 0 .
Substitute y = 0 .
Test the discriminant.
.
.
b 2 4 a c
4 2 4 1 5
16 20
−4
Since the value of the discriminant is negative, there is no solution and so no x- intercept.
Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.
.
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Graph the parabola y = 2 x 2 6 x + 5 .

y : ( 0 , 5 ) ; x : none ;
axis: x = 3 2 ; vertex: ( 3 2 , 1 2 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (3 halves, 1 half). One other point is plotted on the curve at (0, 5). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3 halves.

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Graph the parabola y = −2 x 2 1 .

y : ( 0 , −1 ) ; x : none ;
axis: x = 0 ; vertex: ( 0 , −1 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (0, -1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 0.

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Questions & Answers

where's the answers?
Ed Reply
I don't see where the answers are.
Ed
Cindy and Richard leave their dorm in Charleston at the same time. Cindy rides her bicycle north at a speed of 18 miles per hour. Richard rides his bicycle south at a speed of 14 miles per hour. How long will it take them to be 96 miles apart?
Maddy Reply
3
Christopher
18t+14t=96 32t=96 32/96 3
Christopher
show that a^n-b^2n is divisible by a-b
Florence Reply
What does 3 times your weight right now
Cherokee Reply
Use algebra to combine 39×5 and the half sum of travel of 59+30
Cherokee
What is the segment of 13? Explain
Cherokee
my weight is 49. So 3 times is 147
Cherokee
kg to lbs you goin to convert 2.2 or one if the same unit your going to time your body weight by 3. example if my body weight is 210lb. what would be my weight if I was 3 times as much in kg. that's you do 210 x3 = 630lb. then 630 x 2.2= .... hope this helps
tyler
How to convert grams to pounds?
paul
What is the lcm of 340
Kendra Reply
Yes
Cherokee
How many numbers each equal to y must be taken to make 15xy
Malik Reply
15x
Martin
15x
Asamoah
15x
Hugo
1y
Tom
1y x 15y
Tom
find the equation whose roots are 1 and 2
Adda Reply
(x - 2)(x -1)=0 so equation is x^2-x+2=0
Ranu
I believe it's x^2-3x+2
NerdNamedGerg
because the X's multiply by the -2 and the -1 and than combine like terms
NerdNamedGerg
find the equation whose roots are -1 and 4
Adda
Ans = ×^2-3×+2
Gee
find the equation whose roots are -2 and -1
Adda
(×+1)(×-4) = x^2-3×-4
Gee
Quadratic equations involving factorization
Winner Reply
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Nana
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5x + 1/3= 2x + 1/2
sanam
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sanam
5x - 3x = 1/2 - 1/3 2x = 1/6 x = 1/12
Ranu
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sanam
a trader gains 20 rupees loses 42 rupees and then gains ten rupees Express algebraically the result of his transactions
vinaya Reply
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
Kim is making eight gallons of punch from fruit juice and soda. The fruit juice costs $6.04 per gallon and the soda costs $4.28 per gallon. How much fruit juice and how much soda should she use so that the punch costs $5.71 per gallon?
Mohamed Reply
(a+b)(p+q+r)(b+c)(p+q+r)(c+a) (p+q+r)
muhammad Reply
4x-7y=8 2x-7y=1 what is the answer?
Ramil Reply
x=7/2 & y=6/7
Pbp
x=7/2 & y=6/7 use Elimination
Debra
true
bismark
factoriz e
usman
4x-7y=8 X=7/4y+2 and 2x-7y=1 x=7/2y+1/2
Peggie
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Frank
thanks
Ramil
copy and complete the table. x. 5. 8. 12. then 9x-5. to the 2nd power+4. then 2xto the second power +3x
Sandra Reply
What is c+4=8
Penny Reply
2
Letha
4
Lolita
4
Rich
4
thinking
C+4=8 -4 -4 C =4
thinking
I need to study
Letha
4+4=8
William
During two years in college, a student earned $9,500. The second year, she earned $500 more than twice the amount she earned the first year.
Nicole Reply
9500=500+2x
Debra
9500-500=9000 9000÷2×=4500 X=4500
Debra
X + Y = 9500....... & Y = 500 + 2X so.... X + 500 + 2X = 9500, them X = 3000 & Y = 6500
Pbp
Practice Key Terms 6

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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