Question

Can u help me with this question nd if u can't tell it 2,3

Answer 1

we are given the following equation:

`y=ax^2+bx+c`

We are also given the points:

`\begin{gathered} (-2,-13) \\ (2,3) \\ (4,5) \end{gathered}`

replacing the points:

`\begin{gathered} -13=a(-2)^2+b(-2)+c \\ 3=a(2)^2+b(2)+c \\ 5=a(4)^2+b(4)+c \end{gathered}`

Simplifying:

`\begin{gathered} 4a-2b+c=-13,(1) \\ 4a+2b+c=3,(2) \\ 16a+4b+c=5,(3) \end{gathered}`

We get three equations and three variables. To solve the system we will add equations (1) and (2):

`\begin{gathered} 4a-2b+c+4a+2b+c=-13+3 \\ 8a+2c=-10,(4) \end{gathered}`

Now we multiply equation (2)by -2 and add that to equation (3):

`\begin{gathered} -2(4a+2b+c)+16a+4b+c=-2(3)+5 \\ -8a-4b-2c+16a+4b+c=-6+5 \\ 8a-c=-1,(5) \end{gathered}`

Now we multiply equation(4) by -1 and add that to equation (5):

`\begin{gathered} -(8a+2c)+8a-c=-(-10)-1 \\ -8a-2c+8a-c=10-1 \\ -3c=9 \\ c=(9)/(-3)=-3 \end{gathered}`

Therefore, c = -3. replacing the value of "c" in equation (5):

`\begin{gathered} 8a-(-3)=-1 \\ 8a+3=-1 \end{gathered}`

subtracting 3 to both sides:

`\begin{gathered} 8a=-1-3 \\ 8a=-4 \end{gathered}`

Dividing both sides by 8:

`a=-(4)/(8)=-(1)/(2)`

Now we replace the values of "a" and "c" in equation (1):

`4(-(1)/(2))-2b-3=-13`

Simplifying:

`\begin{gathered} -2-2b-3=-13 \\ -2b-5=-13 \end{gathered}`

Adding 5 to both sides:

`\begin{gathered} -2b=-13+5 \\ -2b=-8 \end{gathered}`

Dividing both sides by -2:

`b=-(8)/(-2)=4`

Therefore, the values of the constants are:

`a=-(1)/(2),b=4,c=-3`