# college pretest

Printable View

• Jun 26th 2006, 03:42 PM
Brooke
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
• Jun 26th 2006, 04:31 PM
Soltras
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.
• Jun 26th 2006, 04:34 PM
topsquark
Quote:

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
• Jun 26th 2006, 04:39 PM
topsquark
Quote:

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
• Jun 26th 2006, 04:44 PM
Soltras
Ahh right, I forgot about $f(x)=a(x-h)^2+k$!
That is much easier than my method.