Question

1. find the equation in standard form of the parabola that passes through the points (0, 6), (-3, 15), and (-6, 6).

Explanation:

Step1: Recall the standard form of a parabola

The standard form of a parabola is \( y = ax^2 + bx + c \). We need to find the values of \( a \), \( b \), and \( c \) using the given points.

Step2: Use the point \((0, 6)\) to find \( c \)

Substitute \( x = 0 \) and \( y = 6 \) into the equation \( y = ax^2 + bx + c \):
\[
6 = a(0)^2 + b(0) + c
\]
Simplifying, we get \( c = 6 \). So the equation becomes \( y = ax^2 + bx + 6 \).

Step3: Use the point \((-3, 15)\) to form an equation

Substitute \( x = -3 \), \( y = 15 \), and \( c = 6 \) into \( y = ax^2 + bx + 6 \):
\[
15 = a(-3)^2 + b(-3) + 6
\]
Simplify the left - hand side and the right - hand side:
\[
15=9a - 3b+6
\]
Subtract 6 from both sides:
\[
9a-3b=15 - 6=9
\]
Divide the entire equation by 3:
\[
3a - b=3\quad(1)
\]

Step4: Use the point \((-6, 6)\) to form an equation

Substitute \( x=-6 \), \( y = 6 \), and \( c = 6 \) into \( y=ax^2+bx + 6 \):
\[
6=a(-6)^2 + b(-6)+6
\]
Simplify the right - hand side:
\[
6 = 36a-6b + 6
\]
Subtract 6 from both sides:
\[
36a-6b=6 - 6=0
\]
Divide the entire equation by 6:
\[
6a - b=0\quad(2)
\]

Step5: Solve the system of equations (1) and (2)

We have the system of equations:
\[

$$\begin{cases} 3a - b=3\\ 6a - b=0 \end{cases}$$

\]
Subtract equation (1) from equation (2):
\[
(6a - b)-(3a - b)=0 - 3
\]
Expand the left - hand side:
\[
6a - b-3a + b=-3
\]
Simplify:
\[
3a=-3
\]
Divide both sides by 3:
\[
a=-1
\]

Step6: Find the value of \( b \)

Substitute \( a=-1 \) into equation (1): \( 3(-1)-b = 3 \)
\[
-3 - b=3
\]
Add 3 to both sides:
\[
-b=3 + 3=6
\]
Multiply both sides by - 1:
\[
b=-6
\]

Step7: Write the equation of the parabola

Now that we have \( a=-1 \), \( b = - 6 \), and \( c = 6 \), the equation of the parabola in standard form is:
\[
y=-x^2-6x + 6
\]

Answer:

The equation of the parabola in standard form is \( y=-x^{2}-6x + 6 \)