Question

What is the equation, in standard form, of a parabola that contains the following points?

(–2, –20), (0, –4), (4, –20)

Answer 1

3 EQUATIONS, 3 UNKNOWNS.

Here's the first one:

\[a(-2)^{2}+b(-2)+c=-20\]

Answer 2

how did u get that?

Answer 3

From the first ordered pair (-2,-20) x=-2, y=-20 and the form of a quadratic is
\[y=ax ^{2}+bx+c\]

Now do tyhe other two points and solve the system.

Answer 4

can u help?

Answer 5

\( (-2, -20)\qquad (0, -4)\qquad (4, -20)\qquad \qquad \large \color{blue}{y=ax^2+bx+c}
\\ \quad \\
\begin{array}{llll}
(-2, -20)\qquad x=-2\quad y=-20\\ \quad \\
-20=a(-2)^2+b(-2)+c\implies &\color{red}{-20=4a-2b+c}\\ \quad \\
(0, -4)\qquad x=0\quad y=-4\\ \quad \\
-4=a(0)^2+b(0)+c\implies &\color{red}{-4=c}\\ \quad \\
(4, -20)\qquad x=4\quad y=-20\\ \quad \\
-20=a(4)^2+b(4)+c\implies &\color{red}{-20=16a+4b+c}
\end{array}\)

Answer 6

you get the system of equations form the \(\bf ax^2+bx+c\) template for a quadratic equation

Answer 7

well \(\bf y= ax^2+bx+c\)

Answer 8

as you can see on the 2nd equation, we only got c = -4
so let us use that in our 2 other equations to reduce them further to a system of equations of 2 variables

4a-2b-4= -20 => 4a - 2b = -16
16a+4b-4 = -20 => 16a+4b = -16

so now you just have a system of equations of 2 variables, and solve that by either substitution or elimination

we already know what "c" is, we just need "a" and "b"

once you get those other 2, plug them back in the template for the quadratic, to get your equation

Answer 9

ok.. thank you!!:)