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Mathematics, 09.02.2021 18:00 lashayreed02

Writing prompt about math Discuss the process of factoring a quadratic function in standard form, y=ax^2+bx+c, as well as using the example y=3x^2+10x+8.

Answers

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Answer from: ComicSans10

c=8

Step-by-step explanation:

Given: y=ax^2+bx+c

To find: c

Solution:

At x=4,y=48:

48=16a+4b+c    (i)

At x=5,y=73:

73=25a+5b+c    (ii)

At x=6,y=104:

104=36a+6b+c   (iii)

On subtracting (i) and (ii),

25=9a+b   (iv)

On subtracting (i) and (iii),

56=20a+2b\\\\28=10a+b    \,\,\,\,(v)

On subtracting (iv) and (v),

a=3

Put a=3 in (iv)

25=9(3)+b\\25=27+b\\25-27=b\\-2=b

Put a=3,b=-2 in (iii)

104=36(3)+6(-2)+c\\104=108-12+c\\104=96+c\\104-96=c\\8=c

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Answer from: angie249

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!

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Answer from: dnarioproctor

Answer is in the photo. I can only upload it to a file hosting service. link below! Good Luck!

https://cutt.ly/wzdvC7R

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Answer from: bookS4169

The quadratic function whose graph contains these points is y=-x^{2}-2x-2

Step-by-step explanation:

We know that a quadratic function is a function of the form y=ax^{2}+bx+c. The first step is use the 3 points given to write 3 equations to find the values of the constants a,b, and c.  

Substitute the points (0,-2), (-5,-17), and (3,-17) into the general form of a quadratic function.

-2=a*0^{2}+b*0+c\\c=-2

-17=a*-5^{2}+b*-5+c\\c=-25a+5b-17

-17=a*3^{2}+b*3+c\\ c=-9a-3b-17

We can solve these system of equations by substitution

Substitute c=-9a-3b-17

-9a-3b-17=25a+5b-17\\-9a-3b-17=-2

Isolate a for the first equation

-9a-3b-17=-25a+5b-17\\a=\frac{b}{2}

Substitute a=\frac{b}{2} into the second equation

-9\left(-\frac{b}{2}\right)-3b-17=-2

Find the value of b

-9\left(-\frac{4b}{17}\right)-3b-17=-2\\ b=-2

Find the value of a

a=\frac{b}{2}\\  a=-1

The solutions to the system of equations are:

b=-2,a=-1,c=-2

So the quadratic function whose graph contains these points is

y=-x^{2}-2x-2

As you can corroborate with the graph of this function.


Aquadratic function is a function of the form y=ax^2+bx+c where a, b, and c are constants. given any
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Answer from: noahwaitsowl357

When variable x is 0 , the value of f(x) gives the y-intercept.

Step-by-step explanation:

When x = 0 , y = c.

c is the y intercept of the function.

ansver
Answer from: garcialopez162017

y=-x²-2x-2

Step-by-step explanation:

Hey, just put x and y value to the statement and you will have 3 of them. Then solve the system :)

From 1st dot we see that c is -2, so a is -1 and b is -2. The result is:

y=-x²-2x-2


Aquadratic function is a function of the form y=ax^2+bx+c where a, b, and c are constants. given any
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Answer from: davidebner2248
A is always true

If a is negative the parabola wil have a minimum value
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Answer from: mauricestepenson791
The general form of the quadratic equation is ax^2 + bx + c = 0.

You are being told, in effect, that 2 points on this curve are (-5,0) and (8,0).

Write out the quadratic equation three times, as follows:

a(-5)^2 + b(-5) + c = 0  and  a(8)^2 + b(8) + c = 0.  Both are = to 0 because y is 0 when x is either -5 or +8.  Also,  y = 90 = a(-2)^2 + b(-2) + c = 0, or 
90 = 4a - 2b + c

Then we have 3 equations in 3 unknowns: 
25a - 5b + c = 0
64a + 8b + c = 0
4a - 2b + c = 90

This is a system of linear equations, and there are various ways of solving such a system, including, but not limited to, substitution, matrices, determinants and the like.  I solved this system on my TI-83 Plus calculator using matrix operations, and found that a = -3, b = -9 and c = 120.

Let me know if you need help in solving such a system.

Thus, the quadratic equation in question is

-3x^2 - 9x + 120 = y

Let's check this out!  Does the point (8,0) satisfy  -3x^2 - 9x + 120 = y  ?

Does -3(8)^2 - 9(8) - 120 = 0 ?  Does -3(64) - 72 + 120 = 0 ?  Unfortunately, NO, which means that there is an arithmetic mistake somewhere in my presentation.  Please read thru this work and ask any questions you may have; we will eventually solve the problem correctly.
ansver
Answer from: aliciaa101

y=(3x+4)(x+2)

Step-by-step explanation:

Once you distibute everything you will get:

3x^2

6x

4x

8

which will give you your intitialequation: y=3x^2+10x+8.  

The first equation you put can not be fcatored.

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