To find the quadratic function that represents the given points, we can start by assuming the quadratic equation has the standard form:

$y = ax^2 + bx + c$

We are given four points from the table: $(-9, 0)$, $(-7, 8)$, $(-5, 8)$, and $(-3, 0)$. Let's use these points to form a system of equations and solve for $a$, $b$, and $c$.

Step 1: Set up the system of equations

For each point $(x, y)$, substitute the values of $x$ and $y$ into the general quadratic equation $y = ax^2 + bx + c$.

• From $(-9, 0)$: $0 = a(-9)^2 + b(-9) + c \quad \Rightarrow \quad 0 = 81a - 9b + c$

• From $(-7, 8)$: $8 = a(-7)^2 + b(-7) + c \quad \Rightarrow \quad 8 = 49a - 7b + c$

• From $(-5, 8)$: $8 = a(-5)^2 + b(-5) + c \quad \Rightarrow \quad 8 = 25a - 5b + c$

• From $(-3, 0)$: $0 = a(-3)^2 + b(-3) + c \quad \Rightarrow \quad 0 = 9a - 3b + c$

Step 2: Solve the system of equations

Now, we have the following system of equations:

1. $81a - 9b + c = 0$
2. $49a - 7b + c = 8$
3. $25a - 5b + c = 8$
4. $9a - 3b + c = 0$

We will solve this system using substitution or elimination. Let's start by subtracting equation 4 from equation 1, and then proceed to solve for $a$, $b$, and $c$.

Let me calculate that for you.The solution to the system of equations is:

$a = -1, \quad b = -12, \quad c = -27$

Therefore, the equation of the quadratic function in standard form is:

$y = -x^2 - 12x - 27$

Would you like a detailed breakdown of how the system was solved?

Here are 5 related questions you might find useful:

1. How can you graph the quadratic function from this equation?
2. How do the vertex and axis of symmetry relate to the equation?
3. What are the roots of this quadratic equation?
4. Can you rewrite this quadratic function in vertex form?
5. How do changes to the coefficients $a$, $b$, and $c$ affect the graph?

Tip: The sign of the coefficient $a$ determines whether the parabola opens upwards (positive $a$) or downwards (negative $a$).