# Afew Questions

• Dec 15th 2007, 03:22 PM
Raj
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 $\displaystyle y=5x-1$ so that the system has (2,9) as its only solution.

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

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

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

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

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

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 $\displaystyle 7(2x+3)^3(x-1) > 0$?

I got $\displaystyle x=\frac{-3}{2},1$
But i think there are others?
• Dec 15th 2007, 03:43 PM
Plato
I must ask you what in the world that posting means?
It seems to be just series of meaningless statements.
• Dec 15th 2007, 03:53 PM
Raj
Was in a rush sort of, i can't believe i forgot my notes at school >.<

I think its more clearer now.
• Dec 15th 2007, 09:26 PM
topsquark
Quote:

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

Answer is $\displaystyle y=2x+5$
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 $\displaystyle y = mx + b$.

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

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

So
$\displaystyle b = 9 - 2m$

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

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

See what you can do with this.

-Dan
• Dec 15th 2007, 09:28 PM
topsquark
Quote:

Originally Posted by Raj
2. The range of the graph $\displaystyle y=\frac{1}{2}(x-12)^2+15$ 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
• Dec 15th 2007, 09:31 PM
topsquark
Quote:

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

Consider the function $\displaystyle y = f(x)$.
When we translate this graph h units to the right the corresponding function is $\displaystyle y = f(x + h)$.

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

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

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

-Dan
• Dec 15th 2007, 09:36 PM
topsquark
Quote:

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

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

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 $\displaystyle b^2-4ac>0$.

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

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

$\displaystyle k<4$

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

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

-Dan
• Dec 15th 2007, 09:42 PM
topsquark
Quote:

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

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

I presume this is really $\displaystyle 7(2x+3)^3(x-1) = 0$?

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
$\displaystyle (x - 1)(x - 2) = 0$
then either
$\displaystyle x - 1 = 0 \implies x = 1$
or
$\displaystyle x - 2 = 0 \implies x = 2$
(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:
$\displaystyle 7(2x+3)^3(x-1) > 0$

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

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

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

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

-Dan