1. ## Afew Questions

Any help with the following questions is appreciated

Linear/Non-Linear Systems of Equations

1. Write an equation to form a system with $y=5x-1$ so that the system has (2,9) as its only solution.

Answer is $y=2x+5$
But how do i get that answer?

2. The range of the graph $y=\frac{1}{2}(x-12)^2+15$ is: ?

1. What is the new equation when $y=(x-2)^2+9$ is translated 5 units left and 3 units down?

1. Determine the values of K so that the graph $f(x)=x^2+4x+k$ crosses the x-axis twice.

$b^2-4ac>0$
$(4)^2-4(1)(k)>0$
$16-4k>0$
$-4k>-16$
$k<4$

Im wondering how to tell if i need a greater/less or equal sign (I think it is > for 2 real roots, < 2 unreal, = roots that are equal) also why it switched on the last step from -4k>-16 to K<4

Polynomial Functions/Equations

1. What are the zeroes of $7(2x+3)^3(x-1) > 0$?

I got $x=\frac{-3}{2},1$
But i think there are others?

2. I must ask you what in the world that posting means?
It seems to be just series of meaningless statements.

3. Was in a rush sort of, i can't believe i forgot my notes at school >.<

I think its more clearer now.

4. Originally Posted by Raj
1. Write an equation to form a system with $y=5x-1$ so that the system has (2,9) as its only solution.

Answer is $y=2x+5$
But how do i get that answer?
I presume that you are looking for another linear function, else there are an infinite number of answers.

So let's assume we have a new equation of the form $y = mx + b$.

So solve the system
$y = 2x + 5$
$y = mx + b$
such that the solution is (2, 9).

I'll start you off:
Put the solution point into the second equation:
$9 = 2m + b$

So
$b = 9 - 2m$

Thus the second equation is
$y = mx + (9 - 2m)$

Now we need to solve the system
$y = 2x + 5$
$y = mx + (9 - 2m)$
for m.

See what you can do with this.

-Dan

5. Originally Posted by Raj
2. The range of the graph $y=\frac{1}{2}(x-12)^2+15$ is: ?
The range of a graph is the set of all possible y values that the function takes. So what y values can this function take?

(Hint: Graph it.)

-Dan

6. Originally Posted by Raj
1. What is the new equation when $y=(x-2)^2+9$ is translated 5 units left and 3 units down?
Consider the function $y = f(x)$.
When we translate this graph h units to the right the corresponding function is $y = f(x + h)$.

When we translate this graph h units to the left the corresponding function is $y = f(x - h)$.

When we translate this graph k units up the corresponding function is $y - k = f(x) \implies y = f(x) + k$.

When we translate this graph k units down the corresponding function is $y + k = f(x) \implies y = f(x) - k$.

-Dan

7. Originally Posted by Raj
1. Determine the values of K so that the graph $f(x)=x^2+4x+k$ crosses the x-axis twice.

$b^2-4ac>0$
$(4)^2-4(1)(k)>0$
$16-4k>0$
$-4k>-16$
$k<4$

Im wondering how to tell if i need a greater/less or equal sign (I think it is > for 2 real roots, < 2 unreal, = roots that are equal) also why it switched on the last step from -4k>-16 to K<4
Note that if the discriminant is equal to 0 then there is only one solution to the quadratic equation. Thus we can't include the equal sign on the line $b^2-4ac>0$.

Whenever you divide both sides of an inequality by a negative number the inequality changes "direction." So
$-4k>-16$

$\frac{-4k}{-4} < \frac{-16}{-4}$

$k<4$

If you have troubles seeing this, consider a simpler example. Solve
$-x > 1$

You can easily get the solution set from the given inequality, and you should be able to see that it is $x < -1$, which is what you get when you divide both sides by -1.

-Dan

8. Originally Posted by Raj
1. What are the zeroes of $7(2x+3)^3(x-1) > 0$?

I got $x=\frac{-3}{2},1$
But i think there are others?
I presume this is really $7(2x+3)^3(x-1) = 0$?

Why should there be other zeros?

The concept behind solving something like this is the following theorem:
If ab = 0 then either a = 0 or b = 0 or both.

So when we a simpler example like
$(x - 1)(x - 2) = 0$
then either
$x - 1 = 0 \implies x = 1$
or
$x - 2 = 0 \implies x = 2$
(and we can't have both terms equal to 0 at the same time.) You've likely seen this process any number of times when solving a quadratic that factors.

In this case we have:
$7(2x+3)^3(x-1) > 0$

$7(2x + 3)(2x + 3)(2x + 3)(x - 1) = 0$

So either
$7 = 0$ <-- Not true!
or
$2x + 3 = 0 \implies x = -\frac{3}{2}$
or
$2x + 3 = 0 \implies x = -\frac{3}{2}$
or
$2x + 3 = 0 \implies x = -\frac{3}{2}$
or
$x - 1 = 0 \implies x = 1$

So we have
$x = -\frac{3}{2}, -\frac{3}{2}, -\frac{3}{2}, 1$

or more simply
$x = -\frac{3}{2}, 1$

-Dan