Question

graph the equation shown below by transforming the given graph of the parent function.
$y = (x + 3)^2 + 4$

Explanation:

Step1: Identify Parent Function

The parent function for \( y=(x + 3)^2+4 \) is \( y = x^2 \), which is a parabola with vertex at \((0,0)\), opening upwards.

Step2: Analyze Horizontal Shift

For the function \( y=(x - h)^2+k \), the horizontal shift is \( h \) units. Here, \( y=(x+3)^2 + 4=(x-(-3))^2+4 \), so there is a horizontal shift of 3 units to the left (since \( h=-3 \)).

Step3: Analyze Vertical Shift

The vertical shift is \( k \) units. Here, \( k = 4 \), so there is a vertical shift of 4 units up.

Step4: Determine New Vertex

The vertex of the parent function \( y=x^2 \) is \((0,0)\). After shifting 3 units left and 4 units up, the new vertex is \((-3,4)\).

Step5: Graph Transformation

Take the graph of \( y = x^2 \), move each point 3 units to the left and 4 units up. For example, the point \((0,0)\) on \( y = x^2 \) moves to \((-3,4)\), the point \((1,1)\) moves to \((-2,5)\), the point \((-1,1)\) moves to \((-4,5)\), etc. The parabola will have the same shape (since there is no vertical stretch or compression, \( a = 1 \)) and will open upwards, with vertex at \((-3,4)\).

Answer:

To graph \( y=(x + 3)^2+4 \), start with the parent function \( y = x^2 \) (a parabola opening up with vertex at \((0,0)\)). Shift the graph 3 units to the left (horizontal shift: \( x\to x + 3 \)) and 4 units up (vertical shift: \( y\to y+4 \)). The resulting parabola has a vertex at \((-3,4)\), opens upwards, and has the same width as \( y = x^2 \).