Questions: F(x)=-1/4(x-1)^2+2 a. What are the coordinates of the vertex? b. Does the graph "open up" or "open down"? c. What is the equation of the axis of symmetry?

Transcript text: $F(x)=-\frac{1}{4}(x-1)^{2}+2$ a. What are the coordinates of the vertex? b. Does the graph "open up" or "open down"? c. What is the equation of the axis of symmetry?

Solution

Solution Steps

Step 1: Converting to Vertex Form

To convert the quadratic function $f(x) = -0.25x^2 + 0x + 2$ to its vertex form, we complete the square. This involves finding values of $h$ and $k$ such that $f(x) = a(x-h)^2 + k$. First, we calculate $h = -\frac0{2a} = -\frac{0}{2_-0.25} = -0$. Then, we find $k$ by substituting $h$ back into the original equation: $k = a_h^2 + b*h + c = 2$. Therefore, the vertex form of the given quadratic function is $f(x) = -0.25(x+0)^2 + 2$.

Step 2: Finding the Vertex

The coordinates of the vertex are directly given by $(h, k) = (-0, 2)$.

Step 3: Determining the Direction of Opening

Since the coefficient $a = -0.25$, the graph opens downward.

Step 4: Identifying the Axis of Symmetry

The axis of symmetry is given by $x = h = -0$.

Final Answer:

The vertex form of the given quadratic function is $f(x) = -0.25(x+0)^2 + 2$, with the vertex at $(-0, 2)$, opening downward, and the axis of symmetry is $x = -0$.

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