Question

graph the parabola and give its vertex, axis of symmetry, x-intercepts, and y-intercept.
$y = \frac{1}{5}x^2 - \frac{8}{5}x + \frac{1}{5}$
a. the x-intercepts are at $x = 0.13, 7.87$ (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
the y-intercept is at $y = \frac{1}{5}$ (type an integer or a simplified fraction)
choose the correct graph of the function, $y = \frac{1}{5}x^2 - \frac{8}{5}x + \frac{1}{5}$, below.
a. graph
b. graph
c. graph
d. graph

Explanation:

Step1: Analyze the parabola equation

The equation of the parabola is \( y = \frac{1}{5}x^2 - \frac{8}{5}x + \frac{1}{5} \). The coefficient of \( x^2 \) is \( \frac{1}{5} \), which is positive, so the parabola opens upwards.

Step2: Find the y - intercept

To find the y - intercept, we set \( x = 0 \) in the equation. Substituting \( x = 0 \) into \( y=\frac{1}{5}x^2-\frac{8}{5}x + \frac{1}{5} \), we get \( y=\frac{1}{5}(0)^2-\frac{8}{5}(0)+\frac{1}{5}=\frac{1}{5} \). So the y - intercept is at \( y = \frac{1}{5} \).

Step3: Find the x - intercepts

To find the x - intercepts, we set \( y = 0 \), so we solve the quadratic equation \( \frac{1}{5}x^2-\frac{8}{5}x+\frac{1}{5}=0 \). Multiply through by 5 to get \( x^2 - 8x + 1=0 \). Using the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) for the quadratic equation \( ax^2+bx + c = 0 \) (here \( a = 1 \), \( b=-8 \), \( c = 1 \)), we have \( x=\frac{8\pm\sqrt{64 - 4(1)(1)}}{2(1)}=\frac{8\pm\sqrt{60}}{2}=\frac{8\pm2\sqrt{15}}{2}=4\pm\sqrt{15}\approx4\pm3.87 \). So \( x\approx4 + 3.87=7.87 \) and \( x\approx4 - 3.87 = 0.13 \), which matches the given x - intercepts.

Step4: Analyze the vertex

The x - coordinate of the vertex of a parabola \( y = ax^2+bx + c \) is \( x=-\frac{b}{2a} \). For \( y=\frac{1}{5}x^2-\frac{8}{5}x+\frac{1}{5} \), \( a=\frac{1}{5} \), \( b =-\frac{8}{5} \). So \( x=-\frac{-\frac{8}{5}}{2\times\frac{1}{5}}=\frac{\frac{8}{5}}{\frac{2}{5}} = 4 \). Substitute \( x = 4 \) into the equation: \( y=\frac{1}{5}(4)^2-\frac{8}{5}(4)+\frac{1}{5}=\frac{16}{5}-\frac{32}{5}+\frac{1}{5}=\frac{16 - 32 + 1}{5}=\frac{-15}{5}=- 3 \). So the vertex is at \( (4,-3) \).

Step5: Match with the graphs

• The parabola opens upwards (since \( a=\frac{1}{5}>0 \)).
• The y - intercept is \( y = \frac{1}{5} \) (close to 0 on the positive y - axis).
• The x - intercepts are approximately 0.13 and 7.87.
• The vertex is at \( (4,-3) \) (in the fourth quadrant, since x = 4>0 and y=-3<0).

Looking at the options:

• Option A: The parabola opens downwards (incorrect, since \( a>0 \)).
• Option B: The y - intercept seems to be at a negative value (incorrect, our y - intercept is positive \( \frac{1}{5} \)).
• Option C: The parabola opens downwards (incorrect, since \( a>0 \)).
• Option D: The parabola opens upwards, has x - intercepts around the correct values, and the vertex is in the correct position.

Answer:

The correct graph is Option D.