10.5 Graphing quadratic equations  (Page 3/15)

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Do these results agree with our graphs? See [link] .

Find the intercepts of a parabola.

To find the intercepts of a parabola with equation $y=a{x}^{2}+bx+c$ :

$\begin{array}{cccc}\hfill {\text{y}}\mathbf{\text{-intercept}}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{\text{x}}\mathbf{\text{-intercepts}}\hfill \\ \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{Let}\phantom{\rule{0.2em}{0ex}}y=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}x.\hfill \end{array}$

Find the intercepts of the parabola $y={x}^{2}-2x-8$ .

Solution

 To find the y -intercept, let $x=0$ and solve for y . When $x=0$ , then $y=-8$ . The y -intercept is the point $\left(0,-8\right)$ . To find the x -intercept, let $y=0$ and solve for x . Solve by factoring.

When $y=0$ , then $x=4\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=-2$ . The x -intercepts are the points $\left(4,0\right)$ and $\left(-2,0\right)$ .

Find the intercepts of the parabola $y={x}^{2}+2x-8.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-8\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4,0\right),\left(2,0\right)$

Find the intercepts of the parabola $y={x}^{2}-4x-12.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-12\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(6,0\right),\left(-2,0\right)$

In this chapter, we have been solving quadratic equations of the form $a{x}^{2}+bx+c=0$ . We solved for $x$ and the results were the solutions to the equation.

We are now looking at quadratic equations in two variables of the form $y=a{x}^{2}+bx+c$ . The graphs of these equations are parabolas. The x -intercepts of the parabolas occur where $y=0$ .

For example:

$\begin{array}{cccc}\hfill \mathbf{\text{Quadratic equation}}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\mathbf{\text{Quadratic equation in two variables}}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}y={x}^{2}-2x-15\hfill \\ \hfill \begin{array}{ccc}\hfill {x}^{2}-2x-15& =\hfill & 0\hfill \\ \hfill \left(x-5\right)\left(x+3\right)& =\hfill & 0\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{let}\phantom{\rule{0.2em}{0ex}}y=0\phantom{\rule{1em}{0ex}}\begin{array}{c}0={x}^{2}-2x-15\hfill \\ 0=\left(x-5\right)\left(x+3\right)\hfill \end{array}\hfill \\ \hfill \begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(-3,0\right)\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}x\text{-intercepts}\hfill \end{array}$

The solutions of the quadratic equation are the $x$ values of the x -intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the equations give the x -intercepts of the graphs, the number of x -intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic equation of the form $a{x}^{2}+bx+c=0$ . Now, we can use the discriminant to tell us how many x -intercepts there are on the graph.

Before you start solving the quadratic equation to find the values of the x -intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Find the intercepts of the parabola $y=5{x}^{2}+x+4$ .

Solution

 To find the y -intercept, let $x=0$ and solve for y . When $x=0$ , then $y=4$ . The y -intercept is the point $\left(0,4\right)$ . To find the x -intercept, let $y=0$ and solve for x . Find the value of the discriminant to predict the number of solutions and so x -intercepts. $\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& -\hfill & 4ac\hfill \\ \hfill {1}^{2}& -\hfill & 4\cdot 5\cdot 4\hfill \\ 1\hfill & -\hfill & 80\hfill \end{array}\hfill \\ \hfill -79\phantom{\rule{0.8em}{0ex}}\hfill \end{array}$ Since the value of the discriminant is negative, there is no real solution to the equation. There are no x -intercepts.

Find the intercepts of the parabola $y=3{x}^{2}+4x+4.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,4\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$

Find the intercepts of the parabola $y={x}^{2}-4x-5.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-5\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\left(-1,0\right)$

Find the intercepts of the parabola $y=4{x}^{2}-12x+9$ .

Solution

 To find the y -intercept, let $x=0$ and solve for y . When $x=0$ , then $y=9$ . The y -intercept is the point $\left(0,9\right)$ . To find the x -intercept, let $y=0$ and solve for x . Find the value of the discriminant to predict the number of solutions and so x -intercepts. $\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& -\hfill & 4ac\hfill \\ \hfill {2}^{2}& -\hfill & 4\cdot 4\cdot 9\hfill \\ \hfill 144& -\hfill & 144\hfill \end{array}\hfill \\ \hfill 0\phantom{\rule{1em}{0ex}}\hfill \end{array}$ Since the value of the discriminant is 0, there is no real solution to the equation. So there is one x -intercept. Solve the equation by factoring the perfect square trinomial. Use the Zero Product Property. Solve for x . When $y=0$ , then $\frac{3}{2}=x.$ The x -intercept is the point $\left(\frac{3}{2},0\right).$

Find the intercepts of the parabola $y=\text{−}{x}^{2}-12x-36.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-36\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-6,0\right)$

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?
3
Christopher
18t+14t=96 32t=96 32/96 3
Christopher
show that a^n-b^2n is divisible by a-b
What does 3 times your weight right now
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
Yes
Cherokee
How many numbers each equal to y must be taken to make 15xy
15x
Martin
15x
Asamoah
15x
Hugo
1y
Tom
1y x 15y
Tom
find the equation whose roots are 1 and 2
(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
Ans = ×^2-3×+2
Gee
find the equation whose roots are -2 and -1
(×+1)(×-4) = x^2-3×-4
Gee
there's a chatting option in the app wow
Nana
That's cool cool
Nana
Nice to meet you all
Nana
you too.
Joan
😃
Nana
Hey you all there are several Free Apps that can really help you to better solve type Equations.
Debra
Debra, which apps specifically. ..?
Nana
am having a course in elementary algebra ,any recommendations ?
samuel
Samuel Addai, me too at ucc elementary algebra as part of my core subjects in science
Nana
me too as part of my core subjects in R M E
Ken
at ABETIFI COLLEGE OF EDUCATION
Ken
ok great. Good to know.
Joan
5x + 1/3= 2x + 1/2
sanam
Plz solve this
sanam
5x - 3x = 1/2 - 1/3 2x = 1/6 x = 1/12
Ranu
Thks ranu
sanam
a trader gains 20 rupees loses 42 rupees and then gains ten rupees Express algebraically the result of his transactions
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 Ok cool answer peggie 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.
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