Question

Describe how you would find the equation of the quadratic function in standard form that passes through the points (-1, 24), (2, -3), and (6, 45)

Answer 1

`y=3x^2-12x+9`

Step-by-step explanation

The standard form of a quadratic function is given by:

`y=ax^2+bx+c`

To find the equation of the function using the given points, we have to find the values of a, b, and c by substituting each of the x and y values from the points into the standard form of the function.

Substituting the values into the function, we will derive three simultaneous equations.

From the first point:

`\begin{gathered} 24=a(-1)^2+b(-1)+c \\ 24=a-b+c \end{gathered}`

From the second point:

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

From the third point:

`\begin{gathered} 45=a(6)^2+b(6)+c \\ 45=36a+6b+c \end{gathered}`

Now, we have three simultaneous equations:

`\begin{gathered} a-b+c=24 \\ 4a+2b+c=-3 \\ 36a+6b+c=45 \end{gathered}`

From the first equation, make a the subject of the formula:

`a=24+b-c`

Now, substitute that into the second and third equations:

`\begin{gathered} \Rightarrow4(24+b-c)+2b+c=-3 \\ 96+4b-4c+2b+c=-3 \\ 96+6b-3c=-3 \\ 6b-3c=-99 \\ 2b-c=-33 \\ \Rightarrow36(24+b-c)+6b+c=45 \\ 864+36b-36c+6b+c=45 \\ 42b-35c=-819 \\ 42b-35c=-819 \\ 6b-5c=-117 \end{gathered}`

Now, we have two new simultaneous equations:

`\begin{gathered} 2b-c=-33 \\ 6b-5c=-117 \end{gathered}`

Make c the subject of the formula in the first equation above:

`c=2b+33`

Substitute that into the second equation and solve for b:

`\begin{gathered} 6b-5(2b+33)=-117 \\ 6b-10b-165=-117 \\ -4b=-117+165=48 \\ \Rightarrow b=(48)/(-4) \\ b=-12 \end{gathered}`

Substitute the value of b into the formula for c:

`\begin{gathered} c=2(-12)+33=-24+33 \\ c=9 \end{gathered}`

Substitute the values of b and c into the formula for a:

`\begin{gathered} a=24+(-12)-9=24-12-9 \\ a=3 \end{gathered}`

Hence, the standard form of the equation of the quadratic equation that passes through the points is:

`y=3x^2-12x+9`