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A friend of mine has a few problems with his math. It's been quite a while since he has done it, and right now he is expected to simply know it all. Therefore, I will post this question on his behalf. I've tried to help him, but I'm not yet this far along with math at school.

$$f(x) = x^2 +10x = -20 + k$$

We started off with $f(x) = 0,5(x+3)^2 + 1$ and $g(x) = 2x + k$. The question is for what $k\in \mathbb R$ does $g(x)$ through the parabola $f(x)$. If somebody could provide us with at least a strategy on how to solve this, my friend and I would be very thankful.

Thanks in advance.

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You start off with the parabola $f(x)=\frac{1}{2}(x+3)^2+1$ and the line $g(x)=2x+k$. First, let's expand $f(x)$ by actually computing the square. What do we get?

$f(x)=\frac{1}{2}x^2+3x+\frac{11}{2}$

Now we will look for their intersection. This happens when...

$f(x)=g(x)$

Now we move everything to one side of the equation to obtain...

$\frac{1}{2}x^2+x+(\frac{11}{2}-k)=0$. Multiply by $2$ to get $x^2+2x+(11-2k)=0$.

Now we use the quadratic formula to find when this equation is true. That is, if $ax^2+bx+c=0$, then $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ So our solutions are...

$x=-1\pm\sqrt{2k-10}$

Since we want the 'stuff' under the square root to be nonnegative, we want...

$2k-10\geq 0$

Then our solution is...

$k \geq 5$. The line intersects the parabola twice when $k>5$ and once when $k=5$.

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