| <<< Note 1425.0 by WFOVX8::SPORBERT "All the little ghoulies are calling your name" >>>
-< Help finding extreme of a parabola? >-
>> I have a question concerning solving a quadratic formula.
>> How do you solve the graph of a quadratic function using the
>> y intercept and an extreme point. Actually I just need to
>> know how to find the extreme point. My school book says
>> something like this:
>> y = ax^2 + bx + c
>> Factor an a out of the two terms containing x.
>> y = a(x^2 + b/ax) + c
>> then complete the square of the terms within the parenthesis.
>> (this is where I get lost...)
>> y = a(x^2 +b/ax + b^2/4a^2) + c - b^2/4a
The idea behind the next step is to make the thing inside the parenteses a
perfect square. Remembering that (w+v)^2 == w^2 + 2*w*v + v^2 and fitting
terms to what we want, we find that if we set w to x and v to b/(2*a), we see
that the equation may be rewritten as
>> y = a(x + b/(2*a))^2 + c - b^2/(4*a)
Now, we want to find the values of x for which y is zero, so we set y to zero
and get
0 = a * (x + b / (2 * a)) ^ 2 + c - b^2 / (4 * a)
In the following, I assume that you want real (as opposed to complex)
solutions.
Moving the non-x terms to the other side and dividing by a (it's not a
quadratic equation unless a is not zero), we get
b^2 / (4 * a^2) - c / a = (x + b / (2 * a)) ^ 2
The right-hand side, being a square, is always at least zero. Thus, if the
left-hand side is negative, we can't solve the equation.
Assuming the equation has solutions, we get that the roots are
sqrt(b^2 - 4*a*c)/(2*a) = x + b/(2*a)
and
- sqrt(b^2 - 4*a*c)/(2*a) = x + b/(2*a)
or, equivalently
x = - b/(2*a) � sqrt(b^2 - 4*a*c)/(2*a)
>> HUH???
I hope this made it easier for you to understand how to solve a quadratic
equation. (If not, I'm a lousy explainer -- please ask again!)
The extremum of the graph is always for the x value -b/(2*a) (trust me!), even
if the equation has no real solutions. The extremum is a maximum if a is
negative and a minimum if a is positive.
>> Is there an easier way to do this???? 8-)
Yes, but only for specific cases. The solution your school book outlined is
the easiest general solution.
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| >> How do you solve the graph of a quadratic function using the
>> y intercept and an extreme point. Actually I just need to
>> know how to find the extreme point.
...
>> y = a(x^2 + b/ax) + c
>> then complete the square of the terms within the parenthesis.
>> (this is where I get lost...)
>> y = a(x^2 +b/ax + b^2/4a^2) + c - b^2/4a
.1 explained how to complete the square well to get to:
y = a(x + b/(2*a))^2 + c - b^2/(4*a)
but then for some reason went on to solve for the roots of x, which is
where the graph crosses the x-axis (the roots are not real numbers iff
the graph doesn't touch or cross the x-axis).
As you said you already know, the y-intercept will be at (0,c). Now you
only need to find where the extreme point is.
Think of it this way.
The basic equation for a parabolic graph that we all learn is y = x�.
This has an extremum at (0,0), and curves upward at what we can call a
"standard" rate. To change the way it "curves", we can change it to
y = rx�
where the graph becomes "taller" or "thinner" if r > 1, becomes "fatter"
if 0 < r < 1, and gets tipped upside-down if r < 0. (I'd rather be
tipped upside-down than squashed flat by r=0. :-)
None of that moves the extreme point at all, of course. To move the whole
graph (and the extreme point with it) to the right by "s", replace "x" by
"x-s" in the equation to get:
y = r(x-s)�
To then move it upward by "t", simply add "t" to the right side of the
equation to get:
y = r(x-s)� + t
This is the easiest form of the equation to figure out the extreme
point from. It's easy to see that the extreme point is where the only
term on the right side that varies, (x-s)�, is at its smallest possible
value of zero, which occurs when x = s. The extreme point is (s,t).
The challenge given to you in your school book is to convert a quadratic
equation from the standard form that you usually see it in,
y = ax� + bx +c,
into the form
y = r(x-s)� + t.
This is why the book went through the steps which have already been
stated to reach:
y = a(x + b/(2a))� + c - b�/(4a)
The only remaining step is to replace the "x + b/(2a)" with "x - (-b/2a)":
y = a(x - (-b/2a))� + c - b�/(4a)
So the extreme point of the parabola defined by y = ax� + bx + c is
( -b b� )
( -- c - -- )
( 2a , 4a )
Jon
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