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The given quadratic function is \( f(x) = x^2 - 10x + 28 \). The vertex of this parabola has already been identified as \( (5, 3) \), and you are tasked with filling in the table for the values of \( f(x) \) at \( x = 4 \) and \( x = 6 \).

Let's find \( f(4) \) and \( f(6) \).

### For \( x = 4 \):
\[
f(4) = 4^2 - 10(4) + 28 = 16 - 40 + 28 = 4
\]

### For \( x = 6 \):
\[
f(6) = 6^2 - 10(6) + 28 = 36 - 60 + 28 = 4
\]

Now, you can complete the table as follows:

\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
4 & 4 \\
5 & 3 \\
6 & 4 \\
\hline
\end{array}
\]

If you have any questions or need further clarification, feel free to ask!

Here are five related questions to expand your understanding:

1. How can you find the axis of symmetry for a quadratic function?
2. What is the significance of the vertex in the context of graphing a quadratic function?
3. How do you determine if the vertex represents a minimum or maximum point?
4. How would the graph change if the coefficient of \( x^2 \) were negative?
5. Can you use a different method (like factoring or completing the square) to solve for the vertex?

**Tip**: The vertex form of a quadratic equation, \( f(x) = a(x-h)^2 + k \), makes it easy to identify the vertex directly.

Find the vertex of f(x) = x^2 - 10x + 28. After this is correctly completed, you will be asked to make a table for the function and graph it. The vertex has been found as (5,3). Complete the table by finding the values of f(x) at x = 4 and x = 6.

Learn how to find the vertex of the quadratic function f(x) = x^2 - 10x + 28 and complete the function table. This problem guides you through determining the vertex, calculating values on either side, and graphing. Suitable for students in Grades 9-10.

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