1. **State the problem:** We are given the graph of a function $f$ which is a straight line passing through points approximately $(-1, 3)$ and $(1, -1)$. We need to find: (a) One value of $x$ such that $f(x) = 3$. (b) The value of $f(1)$. 2. **Find the equation of the line:** The line passes through points $(-1, 3)$ and $(1, -1)$. The slope $m$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{1 - (-1)} = \frac{-4}{2} = -2$$ 3. **Use point-slope form:** Using point $(-1, 3)$, $$y - 3 = -2(x - (-1))$$ $$y - 3 = -2(x + 1)$$ $$y - 3 = -2x - 2$$ $$y = -2x - 2 + 3$$ $$y = -2x + 1$$ So the function is $f(x) = -2x + 1$. 4. **Find $x$ such that $f(x) = 3$:** Set $f(x) = 3$: $$3 = -2x + 1$$ Subtract 1 from both sides: $$3 - 1 = -2x$$ $$2 = -2x$$ Divide both sides by $-2$: $$\frac{2}{\cancel{-2}} = \frac{-2x}{\cancel{-2}}$$ $$x = -1$$ 5. **Find $f(1)$:** $$f(1) = -2(1) + 1 = -2 + 1 = -1$$ **Final answers:** (a) One value of $x$ for which $f(x) = 3$ is $x = -1$. (b) $f(1) = -1$.