Find the minumum point of a function

[Anna55]

Find the minimum point of the function 0.5x^2+x+2.5.
I tired to solve it like this.
0.5x^2+x+2.5=0
However when I plug in this value into the the quadric formula the square root is -4 which one cannot take the square root from. This means the graph does not cut the x-axix. How should you solve it? Thank you in advance!

 

[lookagain]

Find the minimum point of the function 0.5x^2+x+2.5.
I tired to solve it like this.
0.5x^2+x+2.5=0
However when I plug in this value into the the quadric formula the square root is -4
which one cannot take the square root from. This means the graph does not cut the x-axix.
How should you solve it? Thank you in advance!

Anna55,

the square root is not -4. Instead, it is the square root *of* -4. Because the radicand
(quantity that the square root is take of), called the discriminant, is negative, then there
are no real x-intercepts for the corresponding f(x) = 0.5x^2 + x + 2.5. However, you are
not to concern yourself with the discriminant, b^2 - 4ac, for this problem. Instead, use:

\(\displaystyle x = \frac{-b}{2a}**\) for the x-value of the minimum point, and calculate \(\displaystyle f\bigg(\frac{-b}{2a}\bigg)**,\)

which is the y-value of the minimum point.


\(\displaystyle **\)These are for the coordinates of a vertex in general, whether it is a
minimum point or a maximum point.

 

[JeffM]

Find the minimum point of the function 0.5x^2+x+2.5.
I tired to solve it like this.
0.5x^2+x+2.5=0
However when I plug in this value into the the quadric formula the square root is -4
which one cannot take the square root from. This means the graph does not cut the x-axix.
How should you solve it? Thank you in advance!

Anna55,

the square root is not -4. Instead, it is the square root *of* -4. Because the radicand
(quantity that the square root is take of), called the discriminant, is negative, then there
are no real x-intercepts for the corresponding f(x) = 0.5x^2 + x + 2.5. However, you are
not to concern yourself with the discriminant, b^2 - 4ac, for this problem. Instead, use:

\(\displaystyle x = \frac{-b}{2a}**\) for the x-value of the minimum point, and calculate \(\displaystyle f\bigg(\frac{-b}{2a}\bigg)**,\)

which is the y-value of the minimum point.


\(\displaystyle **\)These are for the coordinates of a vertex in general, whether it is a
minimum point or a maximum point.

Anna

The prior response will give you the right answer because the minimum or maximum of F(x) = y = ax^2 + bx + c occurs when x = [(- b) / 2a]. So the minimum value of y is f[(- b) / 2a]. That formula, and others like it, are very easy to derive using differential calculus. You, however, are studying intermediate algebra, and you probably cannot see why that formula is correct.

I really believe that this is a very hard problem to solve for an algebra student. Here is how I recommend that you think about such problems if you do not understand differential calculus. First, sketch a graph of the function to be minimized (this helps in calculus too). This sketch will give you some clues about the approximate answer. You will see in this case that your function has a cup shape (actually a parabola) with a minimum slightly to the left of zero. If you think about it a while, you will realize that if you changed the constant term, the 2.5, for example to 1.5, the graph would be lower everywhere by exactly the same amount, in my example by 1. So WHERE the minimum is found would not change by changing the constant term, just WHAT the minimum is. Are you with me?

Now if you reduced the constant by just the right amount, the reduced function would hit the x-axis in just one point. But the reduced function would equal zero at that point. So you can use the quadratic formula for that reduced function. And it would give you just one root. When you have just one root, the radican equals zero, right? So, the quadratic formula reduces to [(- b) / 2a] for the reduced function, which is the same formula as was given above, but you can get there on your own with algebraic tools and no knowledge of calculus.

Congratulations: you have just solved a calculus problem using only algebra. Step 1: you sketch the graph. Step 2: you think about how to simplify your problem to one that your tools are sufficient to solve and translate that simplified result back into the unsimplified problem. Those two tricks alone will help you in a lot of math.

 

[lookagain]

...the minimum or maximum of \(\displaystyle f(x)\) = y = ax^2 + bx + c occurs when \(\displaystyle x = -b/(2a).\)

So the minimum value of y is \(\displaystyle f(-b/(2a))\).

So, the quadratic formula reduces to \(\displaystyle - b/(2a) ...\)

Stay with the same case of a letter throughout the same problem/discussion, as in "f(x),"
when used to describe the same variable, etc.

You need grouping symbols around this particular denominator, 2a, because of the
Order of Operations. The parentheses around this particular numerator, -b, are
optional.