Where functions meet
- Thread starter jess5085
- Start date
D
Deleted member 4993
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What are your thoughts?
If you are stuck at the beginning tell us and we'll start with the definitions.
Please share your work with us ... even if you know it is wrong
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Hint:
Find the points where y = x+c intersects the curve in terms of 'c'.
Now check for what value of 'c', those points will be in complex domain.
ClaireWu
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- Joined
- Jan 26, 2015
- Messages
- 15
For the line not to meet the parabola, the discriminant b^2-4ac' < 0 for the quadratic equation 2x^2-3x-7=x+c,
rearranging, we have 2x^2-4x-7-c=0 and find values of a,b and c' where a=2,b=-4 and c'=-7-c and solve the discriminant inequality
for range of c
Last edited:
Thanks for your help!
I substituted the values in
(-4)^2 - 4(2)(-7-c)
=16+56+c
=72 +c
72+c < 0
c<-72
But the answer is C is <-9?
D
Deleted member 4993
Guest
But ..
It seems to me that this problem is written incorrectly and thereby clouds the logic needed to solve the problem. It seems to me the problem should be written:
\begin{array}{l}
(eq1)\,\,\,\,\,{y_1} = 2{x^2} - 3x - 7\\
(eq2)\,\,\,\,\,{y_2} = x + c
\end{array}
What we are attempting to find is the value for c which places the line in contact with the parabola at a single point. All values of c less than that value will place the line below the parabola.
Eqs 1 and 2 deliver a distinct value for y1 and y2 for each value of x, what we want is:
\[(eq3)\,\,\,\,\,{y_1} - {y_2} = {\rm{ }}0\]
so that equations 1 and 2 deliver the same value of y for a particular value of x and c.
\begin{array}{l}
(eq4)\,\,\,\,\,\left( {2{x^2} - 3x - 7} \right) - \left( {x + c} \right) = 0\\
(eq5)\,\,\,\,\,2{x^2} - 4x - 7 + c = 0\\
\end{array}
Now we have two unknowns but seemingly only one condition. BUT …we do have a second condition, namely that the quadratic equation must have a single solution (intersect the parabola at one point only), and this can only occur if it's discriminate is equal to zero, i.e.
\begin{array}{l}
(eq6)\,\,\,\,\,{( - 4)^2} - 4(2)( - 7 + c) = 0\,\,\,\,\,\,\,\,\,\, = > \\
(eq7)\,\,\,\,\,c = - 9
\end{array}
From the figure we see that c must be less than -9 to prevent the line y2 from intersecting with the parabola.
It seems to me that this problem is written incorrectly and thereby clouds the logic needed to solve the problem. It seems to me the problem should be written:
\begin{array}{l}
(eq1)\,\,\,\,\,{y_1} = 2{x^2} - 3x - 7\\
(eq2)\,\,\,\,\,{y_2} = x + c
\end{array}
What we are attempting to find is the value for c which places the line in contact with the parabola at a single point. All values of c less than that value will place the line below the parabola.
Eqs 1 and 2 deliver a distinct value for y1 and y2 for each value of x, what we want is:
\[(eq3)\,\,\,\,\,{y_1} - {y_2} = {\rm{ }}0\]
so that equations 1 and 2 deliver the same value of y for a particular value of x and c.
\begin{array}{l}
(eq4)\,\,\,\,\,\left( {2{x^2} - 3x - 7} \right) - \left( {x + c} \right) = 0\\
(eq5)\,\,\,\,\,2{x^2} - 4x - 7 + c = 0\\
\end{array}
Now we have two unknowns but seemingly only one condition. BUT …we do have a second condition, namely that the quadratic equation must have a single solution (intersect the parabola at one point only), and this can only occur if it's discriminate is equal to zero, i.e.
\begin{array}{l}
(eq6)\,\,\,\,\,{( - 4)^2} - 4(2)( - 7 + c) = 0\,\,\,\,\,\,\,\,\,\, = > \\
(eq7)\,\,\,\,\,c = - 9
\end{array}
From the figure we see that c must be less than -9 to prevent the line y2 from intersecting with the parabola.