Question

6 a. standard form
b. vertex form
y = -3(x - 1)^2+3
c. factored form
d. table (include the vertex and at least 2 points on each side of the vertex.)
x y
show the first differences and the second differences.
7 a. standard form
y = -x^2 + 10x - 25
b. vertex form
c. factored form
d. table (include the vertex and at least 2 points on each side of the vertex.)
x y
show the first differences and the second differences.
8 a. standard form
b. vertex form
c. factored form
d. table (include the vertex and at least 2 points on each side of the vertex.)
x y
show the first differences and the second differences.

Explanation:

6.

Step1: Expand vertex - form to standard form

\[

$$\begin{align*} y&=-3(x - 1)^2+3\\ &=-3(x^{2}-2x + 1)+3\\ &=-3x^{2}+6x-3 + 3\\ &=-3x^{2}+6x \end{align*}$$

\]

Step2: Factor the standard - form

\[y=-3x^{2}+6x=-3x(x - 2)\]

Step3: Find the vertex

For the vertex - form \(y=-3(x - 1)^2+3\), the vertex is \((1,3)\).

Step4: Create a table

\(x\)	\(y=-3x^{2}+6x\)	First - difference (\(\Delta y\))	Second - difference (\(\Delta^2y\))
\(1\)	\(3\)	\(-3\)	\(-6\)
\(2\)	\(0\)	\(-9\)	\(-6\)
\(3\)	\(-9\)

Step5: Graph

The parabola \(y=-3x^{2}+6x\) opens downwards (because \(a=-3<0\)), with vertex \((1,3)\) and \(x\) - intercepts at \(x = 0\) and \(x = 2\).

Step1: The given function \(y=-x^{2}+10x - 25\) is already in standard form.

Step2: Convert to vertex - form

\[

$$\begin{align*} y&=-x^{2}+10x - 25\\ &=-(x^{2}-10x + 25)\\ &=-(x - 5)^2+0 \end{align*}$$

\]

Step3: Factor the function

\[y=-x^{2}+10x - 25=-(x - 5)(x - 5)\]

Step4: Find the vertex

For the vertex - form \(y=-(x - 5)^2\), the vertex is \((5,0)\).

Step5: Create a table

\(x\)	\(y=-x^{2}+10x - 25\)	First - difference (\(\Delta y\))	Second - difference (\(\Delta^2y\))
\(4\)	\(-1\)	\(1\)	\(-2\)
\(5\)	\(0\)	\(-1\)	\(-2\)
\(6\)	\(-1\)	\(-3\)	\(-2\)
\(7\)	\(-4\)

Step6: Graph

The parabola \(y=-x^{2}+10x - 25\) opens downwards (because \(a=-1<0\)), with vertex \((5,0)\) and \(x\) - intercept at \(x = 5\).

Since no function is provided in the standard - form for this part, we can't fully complete the steps as above. But if we assume a general quadratic function \(y = ax^{2}+bx + c\).

1. The standard form is \(y=ax^{2}+bx + c\).
2. The vertex - form is \(y=a(x - h)^2+k\), where \(h=-\frac{b}{2a}\) and \(k=y(h)\).
3. To factor, we use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) and then \(y=a(x - x_1)(x - x_2)\) where \(x_1\) and \(x_2\) are the roots.
4. To create a table, we first find the vertex \((h,k)\) and then choose values of \(x\) on either side of \(h\) and calculate \(y\) values. Then we find the first and second differences.
5. To graph, we plot the vertex and the points from the table and draw a parabola.

Answer:

a. \(y=-3x^{2}+6x\)
b. \(y=-3(x - 1)^2+3\)
c. \(y=-3x(x - 2)\)
d. See the table above
e. Graph as described above

7.