# Thread: Quadratics intersecting line and parabola !!!

1. ## Quadratics intersecting line and parabola !!!

Hello im sorry i am not too sure what defines pre calc but i assume this is it. I am having an issue with a particular question which tells me to consider a rule for a straight line as t=mx+5, where m is a real integer. the question asks me that by equating the line t with the parabola y show that x^2+(24-m)x+36=0. There is also he establishment of the two lines at the beginning which i am unsure if it plays into, but have still attached. THANK YOU!

2. ## Re: Quadratics intersecting line and parabola !!!

If you were given the parabola $\displaystyle y= 3(x+ 4)^2- 7= 3(x^2+ 8x+ 16)- 7= 3x^2+ 24x+ 41$ and the line $\displaystyle y= 3x+ 5$. Those will intersect where $\displaystyle y= 3x^2+ 24x+ 41= 3x+ 5$ or $\displaystyle 3x^2+ 21x+ 36= 3(x+ 4)(x+ 3)= 0$ so at (-3, -4) and at (-4, -7). But the line is given, instead by t= 3x+ 5. I have no idea what "t" has to do with "y"!

In the second part, if you were given the more general line, y= mx+ 5. Those will intersect where $\displaystyle y= 3x^2+ 24x+ 41= mx+ 5$
or $\displaystyle 3x^2+ (24- m)x+ 36= 0$. Notice that is NOT the equation given because of the "3" coefficient of $\displaystyle x^2$. And again the problem says t= mx+ 5, not y. I don't know if those are errors or if there is more to this problem that you have not shown.

3. ## Re: Quadratics intersecting line and parabola !!!

Thank you for your response! There is no more relevant context to that equation so I am unsure if it is a mistake or not