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 $\displaystyle (6x-5)(x-2)$.