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

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- Dec 21st 2008, 12:09 PMmagentaritaSmallest Possible y Value
If y = [(4 - x)]^2, what is the smallest possible y value?

- Dec 21st 2008, 12:13 PMMush
- Dec 21st 2008, 12:34 PMmasters
- Dec 21st 2008, 07:10 PMmagentaritaagain...
- Dec 21st 2008, 07:11 PMmagentaritaok.........
- Dec 22nd 2008, 06:23 AMHallsofIvy
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 PMmagentaritaok....
- Dec 23rd 2008, 05:23 AMmasters
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 AMmagentaritaok..I see...
- Dec 23rd 2008, 02:10 PMHallsofIvy
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 PMmagentaritaok.....