To find the quadratic function that goes through the points $(0, 15)$, $(3, 0)$, and $(5, 0)$, we will use the general form of a quadratic equation:

$y = ax^2 + bx + c$

We are given the three points, which will allow us to create a system of equations.

1. For the point (0, 15): When $x = 0$, $y = 15$: $15 = a(0)^2 + b(0) + c \quad \Rightarrow \quad c = 15$

2. For the point (3, 0): When $x = 3$, $y = 0$: $0 = a(3)^2 + b(3) + 15 \quad \Rightarrow \quad 0 = 9a + 3b + 15$ This simplifies to:

\quad \text{(Equation 1)}$$ 3. **For the point (5, 0)**: When $$x = 5$$, $$y = 0$$: $$0 = a(5)^2 + b(5) + 15 \quad \Rightarrow \quad 0 = 25a + 5b + 15$$ This simplifies to: $$25a + 5b = -15 \quad \text{(Equation 2)}$$ Now, we solve this system of equations: ### Step 1: Solve for $$b$$ in terms of $$a$$ From Equation 1: $$9a + 3b = -15 \quad \Rightarrow \quad 3b = -15 - 9a \quad \Rightarrow \quad b = -5 - 3a$$ ### Step 2: Substitute $$b = -5 - 3a$$ into Equation 2 $$25a + 5(-5 - 3a) = -15$$ Simplify: $$25a - 25 - 15a = -15 \quad \Rightarrow \quad 10a - 25 = -15$$ Solve for $$a$$: $$10a = 10 \quad \Rightarrow \quad a = 1$$ ### Step 3: Solve for $$b$$ Substitute $$a = 1$$ into $$b = -5 - 3a$$: $$b = -5 - 3(1) = -8$$ ### Step 4: Write the quadratic function Now that we have $$a = 1$$, $$b = -8$$, and $$c = 15$$, the quadratic function is: $$y = x^2 - 8x + 15$$ This is the quadratic function