To solve this problem, let's determine the x-intercept of the function $f(x)$ and compare it to the other given functions.

Step 1: Find the x-intercept of $f(x)$

The x-intercept occurs when $f(x) = 0$. Using the table, we see:

• For $x = 1$, $f(x) = 1$, and for $x = 5$, $f(x) = -1$.

Using these points, calculate the slope of $f(x)$: $\text{slope} = m = \frac{f(5) - f(1)}{5 - 1} = \frac{-1 - 1}{5 - 1} = \frac{-2}{4} = -\frac{1}{2}.$

Now, the equation of $f(x)$ can be written as: $f(x) = -\frac{1}{2}x + b.$

To find $b$, use the point $(1, 1)$: $1 = -\frac{1}{2}(1) + b \implies b = 1 + \frac{1}{2} = \frac{3}{2}.$

Thus, the equation of $f(x)$ is: $f(x) = -\frac{1}{2}x + \frac{3}{2}.$

Now solve for the x-intercept by setting $f(x) = 0$: $0 = -\frac{1}{2}x + \frac{3}{2} \implies \frac{1}{2}x = \frac{3}{2} \implies x = 3.$

So, the x-intercept of $f(x)$ is $x = 3$.

Step 2: Check the x-intercept of the other functions

1. For $g(x) = -2x + 5$: Set $g(x) = 0$: $0 = -2x + 5 \implies 2x = 5 \implies x = \frac{5}{2}.$ The x-intercept is not $3$, so $g(x)$ is incorrect.

2. For $h(x) = (x - 3)^2$: Set $h(x) = 0$: $(x - 3)^2 = 0 \implies x - 3 = 0 \implies x = 3.$ The x-intercept matches $f(x)$, so $h(x)$ is correct.

3. For $k(x) = (x + 3)^2 + 1$: Set $k(x) = 0$: $(x + 3)^2 + 1 = 0.$ This equation has no real solutions because $(x + 3)^2 \geq 0$ and adding 1 makes it always positive. So, $k(x)$ has no x-intercept and is incorrect.

4. For $m(x) = 2x - 6$: Set $m(x) = 0$: $0 = 2x - 6 \implies 2x = 6 \implies x = 3.$ The x-intercept matches $f(x)$, so $m(x)$ is correct.

Final Answer:

The functions that have the same x-intercept as $f(x)$ are: $\boxed{h(x) \text{ and } m(x)}.$

Let me know if you would like a deeper explanation or have any questions!

Related Questions:

1. How do you determine the slope of a linear function from a table?
2. What is the geometric interpretation of an x-intercept?
3. How do you solve quadratic equations to find their roots?
4. Why does $k(x) = (x + 3)^2 + 1$ have no real x-intercept?
5. Can a quadratic function have two x-intercepts? If so, under what conditions?

Tip:

Always verify the x-intercept of a function by substituting $y = 0$ into its equation!