Question

which of the following is the graph of this quadratic function?
$y = \frac{1}{4}x^2 - 2x + 1$

Explanation:

Step1: Find the vertex of the parabola

For a quadratic function \( y = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( y=\frac{1}{4}x^{2}-2x + 1 \), \( a=\frac{1}{4} \) and \( b=-2 \).
So, \( x =-\frac{-2}{2\times\frac{1}{4}}=\frac{2}{\frac{1}{2}} = 4 \).
To find the y - coordinate, substitute \( x = 4 \) into the function:
\( y=\frac{1}{4}(4)^{2}-2(4)+1=\frac{1}{4}\times16 - 8 + 1=4 - 8 + 1=-3 \). So the vertex is at \( (4,-3) \).

Step2: Find the y - intercept

The y - intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function: \( y=\frac{1}{4}(0)^{2}-2(0)+1 = 1 \). So the y - intercept is \( (0,1) \).

Step3: Analyze the graphs

• The first graph: Check the vertex. The vertex seems to be around \( (4,-3) \) and the y - intercept is \( (0,1) \), which matches our calculations.
• The second graph: The vertex is at a negative x - value, which does not match \( x = 4 \), so we can eliminate it.
• The third graph: The y - intercept is \( (0,5) \), which does not match \( y = 1 \) when \( x = 0 \), so we can eliminate it.

Answer:

The first graph (the one with vertex at (4, - 3) and y - intercept at (0,1))