The vertex form of a quadratic equation is given by

\(\displaystyle \left ( x - h \right )^{2} + k\)

Where the vertex is located at \(\displaystyle \left ( h,k \right )\)

giving us \(\displaystyle \left ( x + 3 \right )^{2} +2\).

To find the x-intercepts of the equation, we set the numerator equal to zero.

\(\displaystyle 0=x^2-9\)

\(\displaystyle 9=x^2\)

\(\displaystyle \sqrt{9}=\sqrt{x^2}\)

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

We can determine the lowest possible value of the expression by finding the \(\displaystyle y\)-coordinate of the vertex of the parabola graphed from the equation \(\displaystyle y=3x^{2}+6x-10\). This is done by rewriting the equation in vertex form.

\(\displaystyle 3x^{2}+6x-10\)

\(\displaystyle 3(x^{2}+2x)-10\)

\(\displaystyle 3(x^{2}+2x+1)-3* 1-10\)

\(\displaystyle 3(x+1)^{2}-13\)

The vertex of the parabola \(\displaystyle y=3(x+1)^{2}-13\) is the point \(\displaystyle (-1, -13)\).

The parabola is concave upward (its quadratic coefficient is positive), so \(\displaystyle -13\) represents the minimum value of \(\displaystyle y\). This is our answer.

Vertex of quadratic equation \(\displaystyle y=ax^{2}+bx+c\) is given by \(\displaystyle (h,k)\).

\(\displaystyle h=-\frac{b}{2a}, k=c-\frac{b^{2}}{4a}\)

For \(\displaystyle y=-2x^{2}+6x-5\),

\(\displaystyle h=-\frac{6}{2\times (-2)}=\frac{3}{2},\)

\(\displaystyle k=-5-\frac{6^{2}}{4\times (-2)}=-5+\frac{9}{2}=-\frac{1}{2}\),

so the coordinate of vertex is \(\displaystyle \left(\frac{3}{2}, -\frac{1}{2}\right)\).

Assume y=0,

\(\displaystyle y=8x^{2}+2x-3=0\)

\(\displaystyle (2x-1)(4x+3)=0\)

\(\displaystyle x=\frac{1}{2}\) , \(\displaystyle x=-\frac{3}{4}\)

In order to find the vertex of a parabola, our first step is to find the x-coordinate of its center. If the equation of a parabola has the following form:

\(\displaystyle f(x)=ax^2+bx+c\)

Then the x-coordinate of its center is given by the following formula:

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

For the parabola described in the problem, a=-2 and b=-12, so our center is at:

\(\displaystyle x_{center}=-\frac{-12}{2(-2)}=-3\)

Now that we know the x-coordinate of the parabola's center, we can simply plug this value into the function to find the y-coordinate of the vertex:

\(\displaystyle f(-3)=-2(-3)^2-12(-3)-23=-5\)

So the vertex of the parabola given in the problem is at the point \(\displaystyle (-3,-5)\)

This is a quadratic function. The \(\displaystyle x\)-coordinate of the vertex of the parabola can be determined using the formula \(\displaystyle x = - \frac{b}{2a}\), setting \(\displaystyle a = 2,b = -5\):

\(\displaystyle x = - \frac{b}{2a} = - \frac{-5}{2 \cdot 2} =\frac{5}{4}\)

Now evaluate the function at \(\displaystyle x = \frac{5}{4}\):

\(\displaystyle f(x) = 2 x^{2} - 5x + 9\)

\(\displaystyle f \left ( \frac{5}{4} \right ) = 2 \left ( \frac{5}{4} \right )^{2} - 5 \left ( \frac{5}{4} \right )+ 9\)

\(\displaystyle = 2 \left (\frac{25}{16} \right ) - 5 \left ( \frac{5}{4} \right )+ 9\)

\(\displaystyle = \frac{25}{8} - \frac{25}{4} + 9\)

\(\displaystyle = \frac{25}{8} - \frac{50}{8} + \frac{72}{8} =\frac{47}{8} =5\frac{7}{8}\)

From the vertex, we know that the equation of the parabola will take the form \(\displaystyle y = a (x - 4) ^{2}+ 1\) for some \(\displaystyle a\) .

To calculate that \(\displaystyle a\), we plug in the values from the other point we are given, \(\displaystyle x = 0, y=-7\), and solve for \(\displaystyle a\):

\(\displaystyle y = a (x - 4)^{2} + 1\)

\(\displaystyle -7= a \cdot (0 - 4)^{2} + 1\)

\(\displaystyle -7= a \cdot (- 4)^{2} + 1\)

\(\displaystyle -7= 16a + 1\)

\(\displaystyle 16a = -8\)

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

Now the equation is \(\displaystyle y = -\frac{1}{2} (x - 4)^{2} + 1\). This is not an answer choice, so we need to rewrite it in some way.

Expand the squared term:

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

Distribute the fraction through the parentheses:

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

Combine like terms:

\(\displaystyle y = -\frac{1}{2} x^{2} +4x - 7\)

To find the max or min of \(\displaystyle y=-5x^2+25x+5\), use the vertex formula and substitute the appropriate coefficients.

\(\displaystyle x=\frac{-b}{2a}=\frac{-25}{2(-5)}= \frac{25}{10}=\frac{5}{2}\)

Since the leading coefficient of \(\displaystyle -5x^2\) is negative, the parabola opens down, which indicates that there will be a maximum.

The answer is:  \(\displaystyle \textup{Max at: }x=\frac{5}{2}\)

To simplify \(\displaystyle 6x^2-17x+10\), determine the factors of the first and last term.

The factor possibilities of \(\displaystyle 6x^2\):

\(\displaystyle x, 2x, 3x, 6x\)

The factor possibilities of \(\displaystyle 10\):

\(\displaystyle 1,2,5,10\)

Determine the signs.  Since there is a positive ending term and a negative middle term, the signs of the binomials must be both negative.  Write the pair of parenthesis.

\(\displaystyle (a -b)(c-d) = ac-ad-bc-bd\)

These factors must be manipulated by trial and error to determine the middle term.

The correct selection is \(\displaystyle x, 6x,2, 5\).

The  answer is \(\displaystyle (6x-5)(x-2)\).