Question

graph the equation.
$y = -4|x| + 3$

Explanation:

Step1: Identify the vertex

The general form of an absolute value function is \( y = a|x - h| + k \), where \((h, k)\) is the vertex. For \( y = -4|x| + 3 \), \( h = 0 \) and \( k = 3 \), so the vertex is \((0, 3)\). Plot this point on the graph.

Step2: Determine the slope for \( x \geq 0 \) and \( x < 0 \)

For \( x \geq 0 \), \( |x| = x \), so the equation becomes \( y = -4x + 3 \). The slope here is \(-4\). For \( x < 0 \), \( |x| = -x \), so the equation becomes \( y = -4(-x) + 3 = 4x + 3 \), and the slope here is \( 4 \).

Step3: Find additional points

For \( x = 1 \) (using \( y = -4x + 3 \)): \( y = -4(1) + 3 = -1 \), so the point is \((1, -1)\). For \( x = -1 \) (using \( y = 4x + 3 \)): \( y = 4(-1) + 3 = -1 \), so the point is \((-1, -1)\). For \( x = 2 \): \( y = -4(2) + 3 = -5 \), point \((2, -5)\). For \( x = -2 \): \( y = 4(-2) + 3 = -5 \), point \((-2, -5)\).

Step4: Draw the graph

Plot the vertex \((0, 3)\), then the points \((1, -1)\), \((-1, -1)\), \((2, -5)\), \((-2, -5)\) etc. Draw two lines: one with slope \(-4\) through \((0, 3)\) and \((1, -1)\) (for \( x \geq 0 \)), and one with slope \(4\) through \((0, 3)\) and \((-1, -1)\) (for \( x < 0 \)). The graph should be a V - shaped graph opening downward with vertex at \((0, 3)\).

Answer:

The graph is a V - shaped absolute value graph with vertex at \((0, 3)\), opening downward, passing through points like \((1, -1)\), \((-1, -1)\), \((2, -5)\), \((-2, -5)\) etc. (To graph it, plot the vertex and the additional points, then draw the two linear segments with slopes \(4\) (for \(x < 0\)) and \(-4\) (for \(x\geq0\)) connecting to the vertex.)