Question

graph the equation shown below by transforming the given graph of the parent function.

$y = x^2 + 3$

Explanation:

Step1: Identify the parent function

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

Step2: Analyze the transformation

The given function is \( y = x^2 + 3 \). For a function of the form \( y = f(x)+k \), if \( k>0 \), the graph of \( y = f(x) \) is shifted up by \( k \) units. Here, \( k = 3 \), so we shift the graph of \( y = x^2 \) up by 3 units.

Step3: Determine the new vertex and points

The vertex of \( y = x^2 \) is \((0,0)\). After shifting up 3 units, the new vertex is \((0, 0 + 3)=(0,3)\). For other points, take a point on \( y = x^2 \), say \((1,1)\), \((- 1,1)\), \((2,4)\), \((-2,4)\). After shifting up 3 units, these points become \((1,1 + 3)=(1,4)\), \((-1,1 + 3)=(-1,4)\), \((2,4 + 3)=(2,7)\), \((-2,4 + 3)=(-2,7)\) etc. Then plot these new points and draw the parabola with vertex at \((0,3)\) opening upwards, same shape as \( y = x^2 \).

Answer:

To graph \( y=x^{2}+3 \), shift the graph of the parent function \( y = x^{2} \) (a parabola with vertex at \((0,0)\)) upward by 3 units. The new vertex is \((0,3)\), and the parabola retains the same shape (opening upwards) as the parent function. Key points like \((1,4)\), \((-1,4)\), \((2,7)\), \((-2,7)\) (derived from shifting points of \( y = x^{2} \) up 3 units) can be plotted to sketch the graph.