1. ## college pretest

for the graph of the quadratic function, identify the direction of the opening and the coordinates of the vertex. f(x)=(x+2)raised to the second -3

A: downward 2,3
B: upward -2,-3
C: downward -2,-3
D: upward 2,3

identify the equation of the a quadractic function with the vertex (0,1/4) that passes through the point (-1,17/4) and opens upward

A: f(x)= 4x raised to the second + 1/4
B: f(x)= -4x raissed to the second - 1/4
C: f(x)= 4x raised to the second - 1/4
D: f(x)= -4x raised to the second - 1/2

2. First question

So you have $f(x)=(x+2)^2-3$

Do you know about derivatives? Taking the derivative will help.
First expand:

$f(x) = x^2 + 4x + 1$

Take the first derivative:

$f'(x) = 2x + 4$

This is a new function describing the original function's slope given an $x$. The vertex of the parabola will have a slope of zero, so set $0 = 2x + 4$ and solve for $x$. This will yield the x-value of the vertex, and you can plug it back into the original function to find its corresponding y.

Take the dervative one more time:

$f''(x) = 2$

This new function gives you the the rate with which the slope of the original function changes. It's useful to determine concavity in a situation like this. As a rule of thumb, if the function is a positive constant, then the parabola opens upward. If it's negative, then it opens downward. In this case, it's positive.

3. Originally Posted by Brooke
for the graph of the quadratic function, identify the direction of the opening and the coordinates of the vertex. f(x)=(x+2)raised to the second -3

A: downward 2,3
B: upward -2,-3
C: downward -2,-3
D: upward 2,3
$f(x) = (x+2)^2 - 3$

This is of the (slightly more) general form: $y = a(x-h)^2+k$

Since a = +1 is positive so we know that the parabola opens upward. The point (h, k) is the vertex, so upon comparing we get (h, k) = (-2, -3).

-Dan

4. Originally Posted by Brooke
identify the equation of the a quadractic function with the vertex (0,1/4) that passes through the point (-1,17/4) and opens upward

A: f(x)= 4x raised to the second + 1/4
B: f(x)= -4x raissed to the second - 1/4
C: f(x)= 4x raised to the second - 1/4
D: f(x)= -4x raised to the second - 1/2
Frankly, I'd do this one by the old "plug 'n' chug" method. (You've got two points on the parabola and 4 equations. Only one of them can be the answer so see if those points fit the parabolas listed in the answer.)

More analytically:
$f(x) = a(x-h)^2 + k$

You know that the vertex (h, k) = (0, 1/4) so
$f(x) = a(x-0)^2 + 1/4 = ax^2+1/4$

Since the other point (-1, 17/4) lies ABOVE this point (the vertex), the parabola must open upward. Thus a is positive.

The only possibility open then, is answer A.

-Dan

5. Ahh right, I forgot about $f(x)=a(x-h)^2+k$!
That is much easier than my method.