Multiply the two through the binomial.

\(\displaystyle y=2(3x+3x^2) = 6x+6x^2 = 6x^2+6x\)

Now that this is in the order of the polynomial \(\displaystyle y=ax^2+bx+c\), we can use the vertex formula.

\(\displaystyle x=-\frac{b}{2a}\)

Substitute the known coefficients.

\(\displaystyle x=-\frac{b}{2a} =-\frac{6}{2(6)}=-\frac{1}{2}\)

The answer is:  \(\displaystyle x=-\frac{1}{2}\)

The polynomial is written in the form of:  \(\displaystyle y=ax^2+bx+c\)

This is the standard form for a parabola.

Write the vertex formula, and substitute the known values:

\(\displaystyle x=-\frac{b}{2a}=-\frac{6}{2(-2)} = \frac{6}{4} =\frac{3}{2}\)

The vertex is at:  \(\displaystyle x=\frac{3}{2}\)

Since the coefficient of \(\displaystyle a\) is negative, the curve will open downward, and will have a maximum.

The answer is:  \(\displaystyle \textup{Max at }x=\frac{3}{2}\)

General parabola equation:

\(\displaystyle ax^2+bx+c=y\)

Vertex formula:

\(\displaystyle h=\frac{-b}{2a}\)

Where \(\displaystyle h\) is the \(\displaystyle x\) value at the vertex.

Combining equations:

\(\displaystyle ax^2-2ahx+c=y\)

Plugging in values for vertex:

\(\displaystyle a(-5)^2-2a(-5)(-5)+c=(-5)\)

Solving for \(\displaystyle c\):

\(\displaystyle c=25a-5\)

Returning to:

\(\displaystyle ax^2+bx+c=y\)

combining equations:

\(\displaystyle ax^2+bx+25a-5=y\)

Plugging in values of given intercept:

\(\displaystyle a*0^2+b*0+25a-5=0\)

Solving for \(\displaystyle a\)

\(\displaystyle a=.2\)

Plugging in value:

\(\displaystyle c=25(.2)-5\)

\(\displaystyle c=0\)

Plugging in values for the vertex:

\(\displaystyle .2(-5)^2+b(-5)+0=(-5)\)

\(\displaystyle 5-5b=-5\)

\(\displaystyle b=2\)

Final equation:

\(\displaystyle .2x^2+2x=y\)

A parabola is a curve that can be represented by a quadratic equation.  The only quadratic here is represented by the function \(\displaystyle f(x) = x^{2} - 9\), while the others represent straight lines, circles, and other curves.

It is not necessary to FOIL the binomial in order to solve for the vertex.  Switch the terms of the quantity \(\displaystyle (3-2x)\), and this equation will be in vertex form:

\(\displaystyle y=a(x-h)^2+k\)

\(\displaystyle y=3(-2x+3)^2-1\)

Set the inner quantity equal to zero.

\(\displaystyle -2x+3 = 0\)

Subtract three on both sides.

\(\displaystyle -2x+3 -3= 0-3\)

\(\displaystyle -2x=-3\)

Divide by negative two on both sides.

\(\displaystyle \frac{-2x}{-2}=\frac{-3}{-2}\)

The location of the vertex is at \(\displaystyle x = \frac{3}{2}\).

To determine the point, substitute the value \(\displaystyle x = \frac{3}{2}\) back to the original equation.

\(\displaystyle y=3(3-2(\frac{3}{2}))^2-1 = 3(3-3)^2-1 = 0-1 = -1\)

The point of the vertex is at:   \(\displaystyle ( \frac{3}{2},-1)\)

Since this parabola opens upward, the point of the vertex will be a minimum.

The answer is:  \(\displaystyle ( \frac{3}{2},-1), \textup{Minimum}\)

Write the vertex formula.

\(\displaystyle x=-\frac{b}{2a}\)

The given equation is already in standard polynomial form.

\(\displaystyle a=-2, b=-1, c=-1\)

Substitute the known values into the formula.

\(\displaystyle x=-\frac{b}{2a}=-\frac{-1}{2(-2)} = -\frac{1}{4}\)

Substitute this value back into the original equation to determine the y value.

\(\displaystyle y=-2(-\frac{1}{4})^2-(-\frac{1}{4})-1\)

Simplify this expression.

\(\displaystyle y=-2(\frac{1}{16})+\frac{1}{4}-1 = -\frac{1}{8}+\frac{1}{4}-1\)

\(\displaystyle y=-\frac{1}{8}+\frac{2}{8}-\frac{8}{8}=-\frac{7}{8}\)

The vertex is located at:  \(\displaystyle (-\frac{1}{4},-\frac{7}{8})\)

A parabola is a curve that is represented by a quadratic function. In this case, the only answer that qualifies is \(\displaystyle 0=3x^2+7-y\).  The other answers represent straight lines, and other types of curves. 

Rewrite the equation in standard polynomial form.

\(\displaystyle y=ax^2+bx+c\)

\(\displaystyle y=-3x^2+5x\)

\(\displaystyle a=-3, b=5, c=0\)

Write the vertex formula and substitute the known coefficients.

\(\displaystyle x=-\frac{b}{2a} = -\frac{5}{2(-3)} = \frac{5}{6}\)

The x-value of the vertex is \(\displaystyle \frac{5}{6}\).

Substitute this value back to the original equation.

\(\displaystyle y=-3(\frac{5}{6})^2+5(\frac{5}{6})\)

\(\displaystyle y=-3(\frac{5}{6})(\frac{5}{6})+5(\frac{5}{6})\)

\(\displaystyle y=-\frac{25}{12}+\frac{25}{6} = -\frac{25}{12}+\frac{25(2)}{6(2)} =-\frac{25}{12}+\frac{50}{12}\)

\(\displaystyle y=\frac{25}{12}\)

The vertex is located at:  \(\displaystyle (\frac{5}{6},\frac{25}{12} )\)

The parabola is provided in the form of \(\displaystyle y=ax^2+bx+c\).

Notice that the variable \(\displaystyle b\) in this equation is zero.

This means that:  

\(\displaystyle x=-\frac{b}{2a} = 0\)

Substitute this value back into the original equation \(\displaystyle y=3x^2-6\) to determine the y-value.

\(\displaystyle y=3(0)^2-6\)

\(\displaystyle y=-6\)

The vertex is located at:  \(\displaystyle (0,-6)\)

A parabola that opens downward has the general formula

\(\displaystyle y = -(x-a)^{2} + b\),

as the negative sign in front of the \(\displaystyle (x-a)\) term makes flips the parabola about the horizontal axis.  

By contrast, a parabola of the form \(\displaystyle x = y^{2}\) rotates about the vertical axis, not the horizontal axis. 

Therefore, \(\displaystyle x = 2(y+3)^{2} -5\) is not the equation for a parabola that opens downward.