Question

use the graph of $y = x^2$ to graph the quadratic function. $f(x) = 6x^2$ use the graphing tool to graph the function. click to enlarge graph

Explanation:

Step1: Recall transformation of quadratic functions

The parent function is \( y = x^2 \). For the function \( f(x)=6x^2 \), we use the transformation rule for vertical stretches. If we have a function \( y = a x^2 \), when \( |a|> 1 \), the graph of \( y=x^2 \) is vertically stretched by a factor of \( |a| \). Here \( a = 6>1 \), so the graph of \( f(x) = 6x^2 \) is a vertical stretch of the graph of \( y=x^2 \) by a factor of 6.

Step2: Identify key points

For the parent function \( y=x^2 \), some key points are \( (0,0) \), \( (1,1) \), \( (- 1,1) \), \( (2,4) \), \( (-2,4) \) etc. For the function \( f(x)=6x^2 \), we apply the transformation to these points.

• For \( x = 0 \), \( f(0)=6\times0^2=0 \), so the point is \( (0,0) \).
• For \( x = 1 \), \( f(1)=6\times1^2 = 6 \), so the point is \( (1,6) \).
• For \( x=- 1 \), \( f(-1)=6\times(-1)^2=6 \), so the point is \( (-1,6) \).
• For \( x = 2 \), \( f(2)=6\times2^2=6\times4 = 24 \), but since the graph grid goes up to \( y = 10 \) and \( y=- 10 \), we can also use the fact that the shape is a parabola opening upwards with vertex at \( (0,0) \) and steeper than \( y = x^2 \) because of the vertical stretch.

Step3: Graph the function

Using the key points \( (0,0) \), \( (1,6) \), \( (-1,6) \) and the knowledge that it's a parabola opening upwards (since the coefficient of \( x^2 \) is positive) with a vertical stretch, we can plot the graph. The vertex is at the origin \((0,0)\), and as \( x \) moves away from 0, the \( y \)-values increase much faster than in \( y = x^2 \).

(Note: Since the problem asks to use the graphing tool, the above steps help in understanding how to plot the graph. The actual graphing would involve plotting the vertex at (0,0) and other points like (1,6), (-1,6) and drawing a parabola opening upwards with a steeper slope than \( y=x^2 \))

Answer:

To graph \( f(x) = 6x^2 \), we vertically stretch the graph of \( y = x^2 \) by a factor of 6. The vertex is at \((0,0)\), and other points like \((1,6)\), \((-1,6)\) are plotted, and a parabola opening upwards (steeper than \( y=x^2 \)) is drawn through these points.