Question

the graph of an absolute value function has a vertex at (-2, 3) and passes through the point (-1, 0). using transformations of the parent function, has the graph been dilated by a scale factor other than 1? explain.

Explanation:

Step1: Recall the vertex form of an absolute value function.

The vertex form of an absolute value function is \( y = a|x - h| + k \), where \((h, k)\) is the vertex. Given the vertex \((-2, 3)\), we substitute \( h = -2 \) and \( k = 3 \) into the equation, so it becomes \( y = a|x - (-2)| + 3 = a|x + 2| + 3 \).

Step2: Substitute the point \((-1, 0)\) into the equation to find \( a \).

We know the function passes through \((-1, 0)\), so we substitute \( x = -1 \) and \( y = 0 \) into \( y = a|x + 2| + 3 \). This gives us \( 0 = a|-1 + 2| + 3 \). Simplifying the absolute value, \( |-1 + 2| = |1| = 1 \), so the equation becomes \( 0 = a(1) + 3 \).

Step3: Solve for \( a \).

From \( 0 = a + 3 \), we subtract 3 from both sides to get \( a = -3 \). Since \( |a| = 3
eq 1 \), the graph has been dilated by a scale factor of 3 (because the absolute value of \( a \) represents the vertical dilation factor; if \( |a| > 1 \), it's a vertical stretch, which is a type of dilation).

Answer:

Yes, the graph has been dilated by a scale factor other than 1 (specifically, a vertical stretch by a scale factor of 3). The vertex form of the absolute value function is \( y = a|x - h| + k \). Substituting the vertex \((-2, 3)\) gives \( y = a|x + 2| + 3 \). Substituting the point \((-1, 0)\) gives \( 0 = a|1| + 3 \), so \( a = -3 \). Since \( |a| = 3
eq 1 \), there is a dilation (vertical stretch) with a scale factor of 3.