# Parabola Question

• June 14th 2009, 05:18 PM
Parabola Question
For the parabola $y = 2x^2-5x+3$

i) State the direction of the parabola.

Up, because 2 is positive?

ii) State whether the y value is a min or a max.

Is this referring to the y value in part iii)?

iii) Find the vertex, then state the min or max value by using $x = -b/2a$

What I'm not getting is the way this question is written. For iii), I got a vertex of (5/4, -1/2). What does it mean by state the min or max value for y? I guess it would be -1/2?
• June 14th 2009, 05:48 PM
VonNemo19
Quote:

For the parabola $y = 2x^2-5x+3$

i) State the direction of the parabola.

Up, because 2 is positive?

ii) State whether the y value is a min or a max.

Is this referring to the y value in part iii)?

iii) Find the vertex, then state the min or max value by using $x = -b/2a$

What I'm not getting is the way this question is written. For iii), I got a vertex of (5/4, -1/2). What does it mean by state the min or max value for y? I guess it would be -1/2?

If it's opening upward, can you think of any way that it's vertex could be a maximum?

A vertex is a min or max of a parabola. (the tip, if you will)
• June 14th 2009, 05:51 PM
Since its opening upward, would the y value be a min? I know that if it was opening downward, it would be a max. I'm just not sure if I'm reading this question correctly. Also, what about part iii)?
• June 14th 2009, 06:16 PM
VonNemo19
Quote:

Since its opening upward, would the y value be a min? I know that if it was opening downward, it would be a max. I'm just not sure if I'm reading this question correctly. Also, what about part iii)?

I'm gonna take you through this step by step

Look at $y=x^2$. This is a simple parabola thet upens upward, with it's vertex (minimum), at the origin.

Now look at $y=(x+1)^2$. Well, this is nothing more than $x^2$, the only difference is that it has slid one unit to the left.

now look at $y=(x+1)^2+3$. again, it's still a parabola that opens upward and now slids to the left one unit, and then up three units.

now look at $y=2(x+1)^2+3$, You guesses it, still the same parabola that we just described, but now it has been "streched" upward by a factor of two.

So, know this, we can put any quadratic equation in this form, and see its graph in our minds eye, without having to physically graph it.

So let's look at your particular equation:

$y=2x^2-5x+3$

We've got to do a little trick here to make it look the same as above. Start by factoring out the 2 from the x terms

$y=2(x^2-\frac{5}{2}x)+3$

Now here's the trick

$2(x^2-\frac{5}{2}x+\frac{25}{16}-\frac{25}{16})+3$

We completed the square but subtracted by the same quantity to maintain equality, now watch. Let's distribute the 2 to the last term inside the parentheses so that we are left with a factorable trinomial square on the inside:

$y=2(x^2-\frac{5}{2}+\frac{25}{16})-\frac{25}{8}+\frac{24}{8}$

Now let's factor that trinomial square

$y=2(x-\frac{5}{4})-\frac{1}{8}$

Now it looks like the one above. So we can see that this is a parabola that opens upward, with vertex (5/4,-1/8), that has been "streched by a factor of two