# question

• Sep 24th 2009, 02:12 AM
mark
question
i'm not actually sure what this question is asking me but i've answered the main part of it. it goes:

a straight line has equation $\displaystyle y = 4x + k$ where k is a constant, and a parabola has equation $\displaystyle y = 3x^2 + 12x + 7$. show that the x coordinate of any points of intersection of the line and the parabola satisfy $\displaystyle 3x^2 + 8x + 7 - k = 0$ hence find the range of values of k for which the line and parabola do not intersect.

i got the second bit and came up with $\displaystyle k < \frac{5}{3}$ which i'm sure is right, but i don't understand what its asking me before it says "hence find the range of values etc"

can someone tell me please?

thanks
• Sep 24th 2009, 02:34 AM
Prove It
Quote:

Originally Posted by mark
i'm not actually sure what this question is asking me but i've answered the main part of it. it goes:

a straight line has equation $\displaystyle y = 4x + k$ where k is a constant, and a parabola has equation $\displaystyle y = 3x^2 + 12x + 7$. show that the x coordinate of any points of intersection of the line and the parabola satisfy $\displaystyle 3x^2 + 8x + 7 - k = 0$ hence find the range of values of k for which the line and parabola do not intersect.

i got the second bit and came up with $\displaystyle k < \frac{5}{3}$ which i'm sure is right, but i don't understand what its asking me before it says "hence find the range of values etc"

can someone tell me please?

thanks

The line and quadratic will intersect where the functions are equal.

So $\displaystyle 4x + k = 3x^2 + 12x + 7$

$\displaystyle 0 = 3x^2 + 8x + 7 - k$.

To show the values of $\displaystyle k$ for which there are not any intersections, we need to remember that the discriminant will be negative.

$\displaystyle \Delta = 8^2 - 4(3)(7 - k) < 0$

$\displaystyle 64 - 84 + 12k < 0$

$\displaystyle -20 + 12k < 0$

$\displaystyle 12k < 20$

$\displaystyle k < \frac{5}{3}$.
• Sep 24th 2009, 02:50 AM
mark
the bit i was looking for an explanation for was- "a straight line has equation $\displaystyle y = 4x + k$ where k is a constant, and a parabola has equation $\displaystyle y = 3x^2 + 12x + 7$ show that the x coordinate of any points of intersection of the line and the parabola satisfy $\displaystyle 3x^2 + 8x + 7 - k = 0$"

did you answer that bit in the first part of your answer?
• Sep 24th 2009, 02:51 AM
Prove It
Yes I did.

The two functions will be equal at the points of intersection, so

$\displaystyle 4x + k = 3x^2 + 12x + 7$

and collecting everything on the one side gives

$\displaystyle 0 = 3x^2 + 8x + 7 - k$

as required.