Question

unit 5 review
score: 55/71 answered: 15/18
question 18
given the parabola $f(x) = 4x^2 - 40x + 36$ find each of the following.
the vertex (write your answer as a point).
the axis of symmetry $x = $
does the parabola open up or down?
\bigcirc opens up
\bigcirc opens down
does the parabola have a relative max or a relative min
\bigcirc max
\bigcirc min

Explanation:

Vertex Calculation

Step1: Recall vertex formula for parabola \( f(x) = ax^2 + bx + c \)

The x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). For the function \( f(x) = 4x^{2}-40x + 36 \), we have \( a = 4 \), \( b=-40 \), and \( c = 36 \).
First, calculate the x - coordinate of the vertex:
\( x=-\frac{-40}{2\times4}=\frac{40}{8} = 5 \)

Step2: Find the y - coordinate of the vertex

Substitute \( x = 5 \) into the function \( f(x) \):
\( f(5)=4\times(5)^{2}-40\times5 + 36 \)
\( f(5)=4\times25-200 + 36 \)
\( f(5)=100-200 + 36=-64 \)
So the vertex is \( (5,-64) \)

Axis of Symmetry Calculation

Step1: Recall the formula for the axis of symmetry

For a parabola \( f(x)=ax^{2}+bx + c \), the axis of symmetry is the vertical line \( x =-\frac{b}{2a} \)
We already calculated \( x =-\frac{-40}{2\times4}=5 \)
So the axis of symmetry is \( x = 5 \)

Direction the Parabola Opens

Step1: Recall the rule for the direction of the parabola

For a parabola \( f(x)=ax^{2}+bx + c \), if \( a>0 \), the parabola opens up; if \( a < 0 \), the parabola opens down.
In the function \( f(x)=4x^{2}-40x + 36 \), \( a = 4>0 \), so the parabola opens up.

Relative Max or Min

Answer:

s:

• The vertex: \( (5,-64) \)
• The Axis of Symmetry \( x = 5 \)
• Does the parabola open up or down? Opens up
• Does the parabola have a relative max or a relative min? Min