# Math Help - Function equations, vetex, etc... Plz help!

1. ## Function equations, vetex, etc... Plz help!

1. Write the equation of the constant function, if possible, which passes throught the point P(4,5).

2. Write the equation of the cubic function, if possible, which passes through the points A(0,1), B(2,9), C(4,12), and D(10,15). Express all coefficients as fractions.

3. Determine the coordinates of the vertex of the parabola
y= -1/4 (x+2)^2 -3

4. Determine the coordinates of the vertex of the parabola
y= 3x^2 -6x +1

5. Determine the coordinates of the point of symmetry for the cubic equation
y= 4x^3 - 6x^2 + 9x -3

6. Write the equation of the quadratic function which has x-intercepts 2, and -1, and a y-intercept of 6.

7. Factor completely in bold R, and write your factorization using exact values (no approximations):
2x^3 + 5x ^2 -1

8. Factor completely in bold C, and write your factorization using exact values (no approximations):
x^4 -2x^3 -6x^2 +22x - 15

9. Write the equation of the parabola depicted in the graph illustrated below:
y-intercepts at (0,4) and (0, -4). Vertex at (2,0). Graph is sideways with two arms expanding to the left, vertex to the right at the x-intercept (2,0).

Thank you so much for your help!

2. Is this a take-home test?

Frankly, if you cannot do the first one, you HAVE been sleeping through class.

Throw us a bone. Please show us the first one. It's an eyeball, definition question. No calculation or manipulation required. Just shout out the answer.

3. ## 1. y=5 .... i know, i'm sorry.

Hi...

Sorry, its not a take home test. I know a constant function is y= c. So the equation should be y= 5. Its just the higher order stuff I have trouble with. My teacher just shows us how to do stuff of the calculator and then expects us to know how to work everything on paper step by step.

Thanks to whoever decides to help..

4. Originally Posted by trae
...

2. Write the equation of the cubic function, if possible, which passes through the points A(0,1), B(2,9), C(4,12), and D(10,15). Express all coefficients as fractions.

...
#1 is OK.

to #2.:

The general equation of a cubic function is:

$f(x)=ax^3+bx^2+cx+d$

If the points belong to the graph of the function the coordinates of the points must satisfy this equation. Plug in the coordinates into the general equation and you'll get a system of simultaneous equations:

$\left|\begin{array}{lcr}0+0+0+d&=&1 \\ 8a+4b+2c+d&=&9 \\ 64a+16b+4c+d &=& 12 \\ 1000a+100b+10c+d&=&15 \end{array}\right.$ .... Solve this system for a, b, c, d:

I've got $\{a, b, c, d\}=\left\{\frac1{20}~,~-\frac{37}{40}~,~\frac{113}{20}~,~1\right\}$

5. Good! Why didn't you demonstrate that in the first place?

Anyway...

#3 is another eyeball problem. (-2,-3) Just look at it. It's in the definition.

Okay, complete the square on #4 and do the same thing.

6. Originally Posted by trae
...

8. Factor completely in bold C, and write your factorization using exact values (no approximations):
x^4 -2x^3 -6x^2 +22x - 15

...
By trial and error you'll find that (x-1) must be a factor. Use synthetic division to factor out this term. Another attempt will give you (x+3) as a factor. Use synthetic divison again. The last quadratic term can only be factored if you use complex numbers.

Thus you have:

$x^4 -2x^3 -6x^2 +22x - 15 = (x-1)(x+3)(x^2-4x+5)$

Calculate now the conjugate complex values to get the last two factors.

7. Hello, trae!

I must assume you know the Factor Theorem.
Here's some help . . .

. . $2x^3 + 5x ^2 -1$
The only possible rational roots are: . $\pm1,\;\pm\frac{1}{2}$
. . We find that $x = -\frac{1}{2}$ is a root . . . Hence, $(2x+1)$ is a factor.

Using long division, we have: . $2x^3 + 5x-1 \:=\:(2x+1)(x^2+2x-1)$

From $x^2+2x-1\:=\:0$, we have: . $x \:=\:\frac{-2\pm\sqrt{8}}{2} \:=\:-1 \pm\sqrt{2}$

. . Hence, $\bigg[x - (-1 + \sqrt{2})\bigg]\text{ and }\bigg[x - (-1 - \sqrt{2})\bigg]$ are factors.

Therefore: . $2x^3 + 5x^2 - 1 \;=\;\boxed{(2x+1)\,(x+1-\sqrt{2})\,(x + 1 + \sqrt{2})}$

8. Factor completely in C; write your factorization using exact values:
. . $x^4 -2x^3 -6x^2 +22x - 15$

The only possible rational roots are: . $\pm1,\:\pm3,\:\pm5,\:\pm15$
. . We find that $x = 1$ is a root . . . Hence, $(x-1)$ is a factor.

Long division: . $x^4-2x^3 - 6x^2 + 22x - 15 \;=\;(x-1)(x^3-x^2-7x+15)$

The possible rational roots of the cubic are: . $\pm1,\:\pm3,\:\pm5,\:\pm15$

. . We find that $x = -3$ is a root . . . Hence, $(x+3)$ is a factor.

Long division: . $x^4 - 2x^3 - 6x^2 + 22x - 15 \;=\;(x-1)(x+3)(x^2-4x + 5)$

From $x^2-4x + 5\:=\:0$ we have: . $x \;=\;\frac{4 \pm\sqrt{-4}}{2} \;=\;2 \pm i$

. . The factors are: . $\bigg[x - (2+i)\bigg]\,\text{ and }\,\bigg[x - (2-i)\bigg]$

Therefore: . $x^4-2x^3 - 6x^2 + 22x - 15 \;=\;\boxed{(x-1)\,(x+3)\,(x-2-i)\,(x-2+i)}$

8. Originally Posted by trae
4. Determine the coordinates of the vertex of the parabola
y= 3x^2 -6x +1
In standard form, a parabola with its axis parallel to the y-axis can be written as $(x-h)^2=4p(y-k)$ or $y=a(x-h)^2+k$ with $a=\frac{1}{4p}$, with the vertex at $(h,k)$ and the focus at $(h,k+p)$. As TKHunny said, completing the square should make identification easy.

Originally Posted by trae
5. Determine the coordinates of the point of symmetry for the cubic equation
y= 4x^3 - 6x^2 + 9x -3
Without using calculus, the only method I could come up with involved some really messy and tedious third-degree equations. There might be a simple way to do it, but I had to look at the inflection point:

First, let us make a conjecture that the point of symmetry of a cubic polynomial is located at its inflection point (sounds reasonable, right?). If we take a general cubic equation, $ax^3 + bx^2 + cx + d = 0,\ a \neq 0$, we have $\frac{d^2y}{dx^2}=6ax+2b=0\\\implies x=-\frac{b}{3a}$, and by testing $\frac{d^2y}{dx^2}$ on the intervals $\left(-\infty,-\frac{b}{3a}\right)$ and $\left(-\frac{b}{3a},\infty\right)$, we see that it does indeed change sign at $x=-\frac{b}{3a}$, so a general cubic function will always have an inflection point at $\left(-\frac{b}{3a}, \frac{2b^3}{27a^2}-\frac{bc}{3a}+d\right)$. Thus, we conjecture that the point of symmetry will also occur at $x=-\frac{b}{3a}$.

Next, we may say that a function $f$ is symmetric with respect to the point $P$ if every point on $f$, $(x, f(x))$, has a corresponding point $(x', f(x'))$ such that $P$ is the midpoint of the line segment joining these two points.

To prove that a cubic function will have a point of symmetry at $x=-\frac{b}{3a}$, consider the points $(x', f(x'))=\left(-\frac{2b}{3a}-x,f\left(-\frac{2b}{3a}-x\right)\right)$ corresponding to $(x, f(x))$. Then, for any $x$, the midpoint of a line segment joining these two points will be:
$\mbox{midpoint}\left((x, f(x)), \left(-\frac{2b}{3a}-x,f\left(-\frac{2b}{3a}-x\right)\right)\right)$
$= \left(\frac{x + \left(-\frac{2b}{3a}-x\right)}{2}, \frac{f(x) + \left(f\left(-\frac{2b}{3a}-x\right)\right)}{2}\right)$
$= \left(-\frac{b}{3a}, \frac{2b^3}{27a^2}-\frac{bc}{3a}+d\right)$
Therefore, the inflection point is indeed the point of symmetry.

Having finally done all that, we can conclude that any cubic function will have a point of symmetry at $x=-\frac{b}{3a}$. Now, you should easily be able to use that formula to find the point of symmetry for your function, $y = 4x^3 - 6x^2 + 9x - 3$: $\left(-\frac{b}{3a}, \frac{2b^3}{27a^2}-\frac{bc}{3a}+d\right) = \left(\frac{1}{2},\frac{1}{2}\right)$

Sorry about the long explanation! When I worked out the problem I generalized it for any cubic, but you could write a much simpler proof if you only considered this particular function.

Originally Posted by trae
6. Write the equation of the quadratic function which has x-intercepts 2, and -1, and a y-intercept of 6.
Originally Posted by trae
9. Write the equation of the parabola depicted in the graph illustrated below:
y-intercepts at (0,4) and (0, -4). Vertex at (2,0). Graph is sideways with two arms expanding to the left, vertex to the right at the x-intercept (2,0).
From the problem description for number 6, the function must contain the following points: $(2, 0),\ (-1, 0),\mbox{ and }(0, 6)$. Take the general form for a quadratic equation, $y=ax^2 + bx + c$ and substitute the given coordinates for $x$ and $y$. You should come up with a system of three linear equations in three variables. Do you see how to solve it now? Number 9 is fairly similar, but remember to interchange the variables since it is sideways.