Question

sketch a graph of: $f(x)=-2|x - 4|+3$

Explanation:

Step1: Identify vertex

The general form of an absolute - value function is $y = a|x - h|+k$, and its vertex is $(h,k)$. For $f(x)=-2|x - 4|+3$, the vertex is $(4,3)$.

Step2: Consider the slope for $x\lt4$

When $x\lt4$, $|x - 4|=-(x - 4)$. So $f(x)=-2\times[-(x - 4)]+3=2x - 8 + 3=2x-5$. The slope of this line is $m = 2$.

Step3: Consider the slope for $x\gt4$

When $x\gt4$, $|x - 4|=x - 4$. So $f(x)=-2(x - 4)+3=-2x + 8+3=-2x + 11$. The slope of this line is $m=-2$.

Step4: Plot points and draw graph

Plot the vertex $(4,3)$. For $x = 3$ (less than 4), $f(3)=2\times3-5 = 1$. For $x = 5$ (greater than 4), $f(5)=-2\times5 + 11 = 1$. Connect the points to form a 'V' - shaped graph opening downwards.

Answer:

Sketch a graph with vertex at $(4,3)$, a line with slope 2 for $x\lt4$ and a line with slope - 2 for $x\gt4$.