# Smallest Possible y Value

• Dec 21st 2008, 12:09 PM
magentarita
Smallest Possible y Value
If y = [(4 - x)]^2, what is the smallest possible y value?
• Dec 21st 2008, 12:13 PM
Mush
Quote:

Originally Posted by magentarita
If y = [(4 - x)]^2, what is the smallest possible y value?

Differentiate and set equal to zero to get maximum/minimum. Solve for x, and then put that value of x back into original equation.

$\displaystyle \frac{dy}{dx} = 2(4-x)(-1) = -2(4-x) = -8+2x=0$
• Dec 21st 2008, 12:34 PM
masters
Quote:

Originally Posted by magentarita
If y = [(4 - x)]^2, what is the smallest possible y value?

Hello Magentarita,

$\displaystyle y=(4-x)^2$

$\displaystyle y=16-8x+x^2$

$\displaystyle y=(x-4)^2$

You looking to find the minimum point of the parabola at vertex (4, 0). That would be the y-coordinate there.
• Dec 21st 2008, 07:10 PM
magentarita
again...
Quote:

Originally Posted by Mush
Differentiate and set equal to zero to get maximum/minimum. Solve for x, and then put that value of x back into original equation.

$\displaystyle \frac{dy}{dx} = 2(4-x)(-1) = -2(4-x) = -8+2x=0$

Why did you use calculus for an algebra 2 question?
• Dec 21st 2008, 07:11 PM
magentarita
ok.........
Quote:

Originally Posted by masters
Hello Magentarita,

$\displaystyle y=(4-x)^2$

$\displaystyle y=16-8x+x^2$

$\displaystyle y=(x-4)^2$

You looking to find the minimum point of the parabola at vertex (4, 0). That would be the y-coordinate there.

So, we basically let y = 0 and simplify, right?
• Dec 22nd 2008, 06:23 AM
HallsofIvy
Quote:

Originally Posted by magentarita
So, we basically let y = 0 and simplify, right?

Simplify what? The problem was just to find the lowest possible value of $\displaystyle y= (4- x)^2$. Because that is squared, y can never be negative but if x= 4, $\displaystyle y= (4-4)^2= 0$.

So, y can't be negative but can be 0. What is the smallest value of y?
• Dec 22nd 2008, 09:03 PM
magentarita
ok....
Quote:

Originally Posted by HallsofIvy
Simplify what? The problem was just to find the lowest possible value of $\displaystyle y= (4- x)^2$. Because that is squared, y can never be negative but if x= 4, $\displaystyle y= (4-4)^2= 0$.

So, y can't be negative but can be 0. What is the smallest value of y?

I thought we let y = 0 like this:

0 = (4 - x)^2

0 = (4 - x) (4 - x)

0 = 16 - 8x + x^2

0 = (4 - x) (4 - x)

0 = 4 - x

-4 = -x

-4/-1 = x

4 = x

This is where the point (4, 0) comes from indicating that the smallest possible y value is 0, right?

• Dec 23rd 2008, 05:23 AM
masters
Quote:

Originally Posted by magentarita
I thought we let y = 0 like this:

0 = (4 - x)^2

0 = (4 - x) (4 - x)

0 = 16 - 8x + x^2

0 = (4 - x) (4 - x)

0 = 4 - x

-4 = -x

-4/-1 = x

4 = x

This is where the point (4, 0) comes from indicating that the smallest possible y value is 0, right?

What you did was solve for the x-intercepts. This is where y=0. This is the smallest value for y only when the parabola is tangent to the x-axis, as in this case. The vertex is on the x-axis at (4, 0)

If the vertex had been below or above the x-axis, setting the equation equal to zero and solving will only find you the zeros (x-intercepts), not the minimum point.

To find the x-coordinate of the minimum/maximum point use

$\displaystyle x=\frac{-b}{2a}$ from $\displaystyle f(x)=ax^2+bx+c$

Then, substitute that into your original equation to find the y-coordinate. Thus, the minimum/maximum point would be

$\displaystyle \left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$
• Dec 23rd 2008, 11:13 AM
magentarita
ok..I see...
Quote:

Originally Posted by masters
What you did was solve for the x-intercepts. This is where y=0. This is the smallest value for y only when the parabola is tangent to the x-axis, as in this case. The vertex is on the x-axis at (4, 0)

If the vertex had been below or above the x-axis, setting the equation equal to zero and solving will only find you the zeros (x-intercepts), not the minimum point.

To find the x-coordinate of the minimum/maximum point use

$\displaystyle x=\frac{-b}{2a}$ from $\displaystyle f(x)=ax^2+bx+c$

Then, substitute that into your original equation to find the y-coordinate. Thus, the minimum/maximum point would be

$\displaystyle \left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$

I get it now.
• Dec 23rd 2008, 02:10 PM
HallsofIvy
Quote:

Originally Posted by magentarita
I thought we let y = 0 like this:

0 = (4 - x)^2

0 = (4 - x) (4 - x)

0 = 16 - 8x + x^2

0 = (4 - x) (4 - x)

Why in the world would you multiply out an expression that was already factored and then factor it again?
In any case, the whole point was not to find the x for which y= 0 but the fact that y can be 0 which should have been obvious.
Quote:

0 = 4 - x

-4 = -x

-4/-1 = x

4 = x

This is where the point (4, 0) comes from indicating that the smallest possible y value is 0, right?

• Dec 23rd 2008, 08:50 PM
magentarita
ok.....
Quote:

Originally Posted by HallsofIvy
Why in the world would you multiply out an expression that was already factored and then factor it again?
In any case, the whole point was not to find the x for which y= 0 but the fact that y can be 0 which should have been obvious.

I thank you.