Function Problem

scrible · #1 $Old$ November 15th, 2009, 06:15 AM

I have this function that is just beating me. According to the text it ask to find the value of $k(=/ 1)$ such that the quadratic function of $x$

$k(x+2)^2 -(x-1)(x-2)$
is equal to zero for only one value of x.
Find also(a) the range of values of k for which the function possesses a minimum value, (b) the range of values of k for which the value of the function never exceeds 12.5.
Sketch the graph of the function for $k=\frac{1}{2}$ and for $k=2\frac{1}{2}$ . Can some one help me to solve this problem. Thank you in advance.

ukorov · #2 $Old$ November 15th, 2009, 06:34 AM

what is the meaning of k (= /1) ?

scrible · #3 $Old$ November 15th, 2009, 07:11 AM

The meaning of (k=\1) not equal to one. I can't find a simbol in the math program for not equal to. Is there one?

HallsofIvy · #4 $Old$ November 15th, 2009, 08:41 AM

Quote:

Originally Posted by scrible $View Post$

I have this function that is just beating me. According to the text it ask to find the value of $k(=/ 1)$ such that the quadratic function of $x$

$k(x+2)^2 -(x-1)(x-2)$
is equal to zero for only one value of x.
Find also(a) the range of values of k for which the function possesses a minimum value, (b) the range of values of k for which the value of the function never exceeds 12.5.
Sketch the graph of the function for $k=\frac{1}{2}$ and for $k=2\frac{1}{2}$ . Can some one help me to solve this problem. Thank you in advance.

$k(x+2)^2- (x-1)(x-2)= k(x^2+ 4x+ 4)- (x^2-3x+ 2)= (k-1)x^2+ (4k+3)x+ 4k-2$ . $(k-1)x^2+ (4k+3)x+ 4k-2= 0$ will have a single root if the "discriminant" $b^2- 4ac= (4k+3)^2- 4(k-1)((4k-2)= 0$ . Solve that quadratic equation for k.

For b, complete the square in $(k-1)x^2+ (4k+1)x+ 4k-2$ to determine the vertex of the graph in terms of k then take k so the "y-value" of the vertex is never greater than 12.5. Of course, k-1 will have to be negative in order that that have a maximum.

scrible · #5 $Old$ November 15th, 2009, 01:44 PM

Quote:

Originally Posted by HallsofIvy $View Post$

$k(x+2)^2- (x-1)(x-2)= k(x^2+ 4x+ 4)- (x^2-3x+ 2)= (k-1)x^2+ (4k+3)x+ 4k-2$ . $(k-1)x^2+ (4k+3)x+ 4k-2= 0$ will have a single root if the "discriminant" $b^2- 4ac= (4k+3)^2- 4(k-1)((4k-2)= 0$ . Solve that quadratic equation for k.

For b, complete the square in $(k-1)x^2+ (4k+1)x+ 4k-2$ to determine the vertex of the graph in terms of k then take k so the "y-value" of the vertex is never greater than 12.5. Of course, k-1 will have to be negative in order that that have a maximum.

Thank you very much. I don't know how you got $(k-1)x^2+ (4k+3)x+ 4k-2$ . $(k-1)x^2+ (4k+3)x+ 4k-2= 0$ . Is there a way you can break it down for me?