Question

graph the parabola and give its vertex, axis of symmetry, x-intercepts, and y-intercept
$y = \frac{1}{3}x^2 - \frac{4}{3}x + \frac{1}{3}$
the vertex is (2, -1). (type an ordered pair.)
the axis of symmetry is $x = 2$. (type an equation.)
select the correct choice below and fill in any answer boxes within your choice.
a. the x-intercepts are at $x = \square$. (type an integer or decimal rounded to two decimal places as needed. use a comma to separate answers as needed.)
b. there is no x-intercept

Explanation:

Step1: Set \( y = 0 \)

To find the x - intercepts, we set \( y = 0 \) in the equation \( y=\frac{1}{3}x^{2}-\frac{4}{3}x + \frac{1}{3} \). So we have the equation \( 0=\frac{1}{3}x^{2}-\frac{4}{3}x+\frac{1}{3} \).

Step2: Multiply through by 3

Multiply each term in the equation \( 0=\frac{1}{3}x^{2}-\frac{4}{3}x+\frac{1}{3} \) by 3 to clear the fractions. We get \( 0 = x^{2}-4x + 1 \).

Step3: Use quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \) (here \( a = 1 \), \( b=-4 \), \( c = 1 \)), the quadratic formula is \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-4)^{2}-4\times1\times1=16 - 4=12 \).
Then, \( x=\frac{4\pm\sqrt{12}}{2}=\frac{4\pm2\sqrt{3}}{2}=2\pm\sqrt{3} \).
\( \sqrt{3}\approx1.73 \), so \( x = 2 + 1.73=3.73 \) and \( x=2 - 1.73 = 0.27 \).

Answer:

A. The x - intercepts are at \( x = 0.27, 3.73 \)