Question

graph the function $f(x) = \frac{1}{4}x^2 + 5$. plot the vertex. then plot another point on the parabola. if you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.

Explanation:

Step1: Identify the vertex form

The function \( f(x) = \frac{1}{4}x^2 + 5 \) is in the form \( f(x)=ax^2 + k \), where \( a=\frac{1}{4} \) and \( k = 5 \). For a parabola in the form \( y=ax^2 + k \), the vertex is at \( (0,k) \). So the vertex is \( (0,5) \).

Step2: Find another point

Let's choose \( x = 4 \) (we can choose any \( x \)-value, 4 is a good choice because when we square it, it will make the calculation with \( \frac{1}{4} \) easy). Substitute \( x = 4 \) into the function:
\( f(4)=\frac{1}{4}(4)^2+5 \)
First, calculate \( (4)^2 = 16 \), then \( \frac{1}{4}\times16 = 4 \), then \( 4 + 5=9 \). So the point \( (4,9) \) is on the parabola.

Step3: Graph the parabola

• Plot the vertex \( (0,5) \).
• Plot the point \( (4,9) \). Since the parabola is symmetric about the \( y \)-axis (because there is no \( x \)-term, so the axis of symmetry is \( x = 0 \)), the point \( (- 4,9) \) is also on the parabola. Then draw a smooth curve through these points to form the parabola.

(Note: Since the problem asks to plot the points, the key points to plot are the vertex \( (0,5) \) and another point like \( (4,9) \) or \( (-4,9) \))

Answer:

The vertex of the parabola \( f(x)=\frac{1}{4}x^2 + 5 \) is \( (0,5) \), and another point on the parabola is \( (4,9) \) (or \( (-4,9) \)). To graph the function, plot the vertex \( (0,5) \) and the point \( (4,9) \) (or \( (-4,9) \)) and draw a smooth parabola symmetric about the \( y \)-axis.