Mathematics, 04.11.2020 06:20 alliemmaddox

Find a function of the form y=ax^2+bx+c whose graph passes through (-11,2),(-9,-2)and (-5,14). (Vertex form also?) please help

Find a function of the form y=ax^2+bx+c whose graph passes through (-11,2),(-9,-2)and (-5,14). (Ver

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Answer from: Jasoncookies23

y=x^2+18x+79

y=(x+9)^2-2

Step-by-step explanation:

We want to find a function of the form:

y=ax^2+bx+c

We know that it passes through the three points: (-11, 2); (-9, -2); and (-5,14).

In other words, if we substitute -11 for x, we should get 2 for y. So:

(2)=a(-11)^2+b(-11)+c

Simplify:

2=121a-11b+c

We will do it to the two other points as well. So, for (-9, -2):

$\begin{aligned}(-2)&=a(-9)^2+b(-9)+c\\\Rightarrow-2&=81a-9b+c\end{aligned}$

And similarly:

$\begin{aligned}(14)&=a(-5)^2+b(-5)+c\\\Rightarrow 14&=25a-5b+c\end{aligned}$

So, to find our function, we will need to determine the values of a, b, and c.

We essentially have a triple system of equations:

$\begin{cases}2=121a-11b+c\\-2=81a-9b+c\\14=25a-5b+c\end{cases}$

To approach this, we can first whittle it down only using two variables.

So, let's use the First and Second Equation. Let's remove the variable b. To do so, we can use elimination. We can multiply the First Equation by 9 and the Second Equation by -11. This will yield:

$\begin{cases}{\begin{aligned}9(2)&=9(121a-11b+c)\\-11(-2)&=-11(81a-9b+c)\end{cases}}$

Distribute:

$\begin{cases}18=1089a-99b+9c\\22=-891a+99b-11c\end{cases}$

Now, we can add the two equations together:

(18+22)=(1089a-891a)+(-99b+99b)+(9c-11c)

Simplify:

40=198a-2c

Now, let's do the same using the First and Third Equations. We want to cancel the variable b. So, let's multiply the First Equation by 5 and the Third Equation by -11. So:

$\begin{cases}{\begin{aligned}5(2)=5(121a-11b+c)\\-11(14)=-11(25a-5b+c)\end{cases}$

Simplify:

$\begin{cases}10=605a-55b+5c\\-154=-275a+55b-11c\end{cases}$

Now, let's add the two equations together:

(10-154)=(605a-275a)+(-55b+55b)+(5c-11c)

Simplify:

-144=330a-6c

Therefore, we now have the two equations:

$\begin{cases}40=198a-2c\\-144=330a-6c\end$

Let's cancel the c. So, multiply the First Equation by -3. We don't have to do anything special to the second. So:

-3(40)=-3(198a-2c)

Multiply:

-120=-594a+6c

Now, add it to the Second Equation:

(-120+-144)=(-594a+330a)+(6c-6c)

Add:

-264=-264a

Divide both sides by -264. So, the value of a is:

a=1

Now, we can use either of the two equations above to obtain c. Let's use the first one. So:

40=198a-2c

Substitute 1 for a:

40=198(1)-2c

Solve for c. Subtract 198 from both sides and divide by -2:

$\begin{aligned}40&=198-2c\\ -158&=-2c \\79 &=c \end{aligned}$

So, the value of c is 79.

Finally, we can find b. We can use any of the three original equations. Let's use the First Equation. So:

2=121a-11b+c

Substitute 1 for a and 79 for c and determine the value of b:

$\begin{aligned}2&=121(1)-11b+(79)\\2&=121+79-11b\\2&=200-11b\\-198&=-11b\\18&=b\end{cases}$

Therefore, the value of b is 18.

So, our equation is:

y=ax^2+bx+c

Substitute in the values:

y=(1)x^2+(18)x+(79)

Simplify:

y=x^2+18x+79

Now, let's put this into vertex form. To do so, we will need to complete the square. First, let's group the first two terms together:

y=(x^2+18x)+79

To complete the square, we will divide the b term by 2 and then square it.

b is 18. 18/2 is 9 and 9² is 81. Therefore, we will add 81 into our parentheses:

y=(x^2+18x+81)+79

Since we added 81, we must also subtract 81. So:

y=(x^2+18x+81)+79-81

Subtract:

y=(x^2+18x+81)-2

The grouped terms are a perfect square trinomial. Factor:

y=(x+9)^2-2

And this is in vertex form.

And we are done!

Answer from: Quest

no not all of them.

step-by-step explanation:

some of them have a square base.

Answer from: Quest

0.02040816326

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Find a function of the form y=ax^2+bx+c whose graph passes through (-11,2),(-9,-2)and (-5,14). (Vert...

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