Recall that a value of x for which
f(x) = 0
is called a zero of the quadratic function. Set the function equal to zero to be able to solve for x. Set the given function equal to zero by substituting 0 for f(x).
f(x)     =     5(x − 2)2 − 45
write an answer in an equation where it =     5(x − 2)2 − 45

To find the zero of the given quadratic function, we need to set \( f(x) = 0 \) and then solve for \( x \). The given function is:

\[
f(x) = 5(x - 2)^2 - 45
\]

Now, set \( f(x) = 0 \):

\[
0 = 5(x - 2)^2 - 45
\]

Next, solve for \( x \):

1. Add 45 to both sides:

\[
45 = 5(x - 2)^2
\]

2. Divide both sides by 5:

\[
9 = (x - 2)^2
\]

3. Take the square root of both sides:

\[
\sqrt{9} = |x - 2|
\]

This gives two solutions:

\[
x - 2 = 3 \quad \text{or} \quad x - 2 = -3
\]

4. Solve each equation:

\[
x = 3 + 2 = 5
\]
\[
x = -3 + 2 = -1
\]

Thus, the solutions (or the zeros) are:

\[
x = 5 \quad \text{and} \quad x = -1
\]

Would you like further details on the steps? Here are some follow-up questions to expand on this concept:

1. What is the geometric significance of the zeros in a quadratic function?
2. How would the graph of this quadratic function look based on the values of the zeros?
3. Can we apply the quadratic formula to this function even though it's already factored?
4. What is the role of the leading coefficient (5) in determining the shape of the parabola?
5. How would you check if your solutions for \( x \) are correct?

**Tip**: When solving a quadratic equation, always check for both positive and negative roots when taking square roots.

Recall that a value of x for which f(x) = 0 is called a zero of the quadratic function. Set the function equal to zero to be able to solve for x. Set the given function equal to zero by substituting 0 for f(x). f(x) = 5(x − 2)^2 − 45.

Learn how to solve the quadratic function 5(x - 2)^2 - 45 to find its zeros. This problem demonstrates how to use algebraic techniques, including setting the function equal to zero and applying the square root property. Detailed steps make this ideal for students in Grades 8-10.

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