Question

question
find $f(2)$.
$h(x) = -x + 4$

$x$	$g(x)$
$-1$	$-9$
$-8$	$3$
$-6$	$5$
$6$	$-4$
$-2$	$3$

Explanation:

Step1: Identify the graph type

The graph of \( f(x) \) is a parabola (quadratic function). To find \( f(2) \), we look at the \( x \)-value of 2 on the graph and determine the corresponding \( y \)-value.

Step2: Locate \( x = 2 \) on the graph

On the coordinate plane, when \( x = 2 \), we find the point on the parabola. From the graph, when \( x = 2 \), we can see that the \( y \)-value (which is \( f(2) \)) can be calculated or observed. Alternatively, we can assume the parabola has a vertex form or standard form, but from the graph's grid, each square is 1 unit. Looking at the graph, when \( x = 2 \), the \( y \)-coordinate is 5? Wait, no, let's check again. Wait, the parabola: let's see the vertex. The vertex seems to be at \( x = -1 \), \( y = -4 \)? Wait, no, the graph: when \( x = 0 \), \( y = -3 \)? Wait, maybe I misread. Wait, the graph of \( f(x) \): let's check the coordinates. Wait, the parabola passes through \( x = 1 \), what's \( y \)? Wait, maybe the function is \( f(x) = x^2 - 3 \)? No, when \( x = 2 \), \( x^2 = 4 \), \( 4 - 3 = 1 \)? No. Wait, maybe the graph is \( f(x) = x^2 - 3 \)? Wait, no, let's look at the grid. Each horizontal and vertical line is 1 unit. So when \( x = 2 \), moving up from \( x = 2 \) on the \( x \)-axis, the point on the parabola: let's see, the parabola at \( x = 2 \), the \( y \)-value. Wait, maybe the correct way is to see that the parabola has a vertex at \( (-1, -4) \), so the equation is \( f(x) = (x + 1)^2 - 4 \). Let's expand that: \( f(x) = x^2 + 2x + 1 - 4 = x^2 + 2x - 3 \). Now, plug in \( x = 2 \): \( f(2) = (2)^2 + 2(2) - 3 = 4 + 4 - 3 = 5 \)? Wait, no, 4 + 4 is 8, 8 - 3 is 5? Wait, but let's check the graph. When \( x = 2 \), the \( y \)-coordinate: looking at the graph, at \( x = 2 \), the point is at \( y = 5 \)? Wait, maybe I made a mistake. Wait, the graph: when \( x = 1 \), \( y = 1 + 2 - 3 = 0 \), so the parabola crosses the \( x \)-axis at \( x = 1 \) (since \( f(1) = 0 \)). Then at \( x = 2 \), \( f(2) = 4 + 4 - 3 = 5 \). Yes, that makes sense. So \( f(2) = 5 \). Wait, but let's confirm with the graph. The graph at \( x = 2 \), the \( y \)-value is 5. So that's \( f(2) = 5 \).

Wait, maybe another way: the graph of \( f(x) \) is a parabola. To find \( f(2) \), we go to \( x = 2 \) on the \( x \)-axis, then find the corresponding \( y \)-value on the graph. From the graph, when \( x = 2 \), the \( y \)-value is 5. So \( f(2) = 5 \).

Answer:

\( \boxed{5} \)