For $x$ and $k$ real numbers, determine the values of $k$ for which the graphs of $f(x)=x^2-4$ and $g(x)=2|x|+k$ do not intersect

Question

For $x$ and $k$ real numbers, determine the values of $k$ for which the graphs of $f(x)=x^2-4$ and $g(x)=2|x|+k$ do not intersect.

For $x$ and $k$ real numbers, determine the values of $k$ for which the graphs of $f(x)=x^2-4$ and $g(x)=2|x|+k$ have exactly 2 intersections.

I graphed them on Desmos and found that when k is smaller than -5 they don’t have any intersections and when k is greater or equal to -5 they’ll have two intersections, but I really have no idea what to begin with when it comes to handwriting.

I graphed them on Desmos and found that when k is smaller than -5 they don’t have any intersections and when k is greater or equal to -5 they’ll have two intersections, but I really have no idea what to begin with when it comes to handwriting. — ndjjxhsjak, Sep 12 '19 at 21:47

Quanto · Answer 1 · 2019-09-14T13:04:04.277

3

Set the following equations from $f(x)=g(x)$,

$$ x^2 -4 = 2x+k$$

$$ x^2 -4 =- 2x+k$$

Or,

$$ x^2 -2x -(4+k)= 0$$

$$ x^2 +2x -(4+k)= 0$$

Their discriminant is

$$\Delta =b^2-4ac = 20+4k$$

If they do not intersect, meaning above equations have no solution, we set $\Delta <0$. So, $k <-5$.

If they have two intersections, meaning above equations have one solution each, we set $\Delta =0$. So, $k=-5$

edited Sep 14 '19 at 13:04

answered Sep 12 '19 at 21:54

Quanto

97,352

In the last sentence, there's something wrong. Either you mean "one intersection" or you mean $\Delta>0.$ – saulspatz Sep 12 '19 at 21:59
I do mean two intersections, one with each line – Quanto Sep 12 '19 at 22:00
Sorry but how did you get 20+4k? – ndjjxhsjak Sep 12 '19 at 22:05
From the quadratic equations, $a=1$, $b=\pm 2$, and $c=-(4+k)$. Then, calculate $b^2-4ac$. – Quanto Sep 12 '19 at 22:08

score 2 · Answer 2 · answered Sep 12 '19 at 21:55

If the 2 functions don't have an intersection, then we must have that $f(x) \ne g(x)$ for all $x$, i.e. $x^2-4 \ne 2|x|+k$ which gives that $x^2 -2|x| -4-k \ne 0$.

Do you know what to do with the absolute value? Try to take 2 cases, if $|x|= x$ and if $|x|=-x$

And find the discriminant of this quadratic equation, when is it negative so that it doesn't have roots?

user0102 · Answer 3 · 2019-09-13T00:11:32.367

You may also think about this problem geometrically. The image of the function $f(x) = x^{2} - 4$ is a parabola and the image of the function $g(x) = 2|x| + k$ are half-lines centered at the origin lying in the first and second quadrant shifted by the constant $k$. Thus their graphs intersect exactly twice when the graph of $g$ tangents the graph of $f$. More precisely, if $(a,b)$ is the right tangent point, the following system has to have exactly one solution: \begin{align*} \begin{cases} 2a + k = b\\ a^{2} - 4 = b \end{cases} \Longrightarrow a^{2} - 2a - k - 4 = 0 \Longrightarrow \Delta = 4 + 4k + 16 = 0 \Longrightarrow k = -5 \end{align*}

Due to the symmetry of $g$'s and $f$'s graphs, when $k = -5$, $\text{Gr}(f)\cap\text{Gr}(g) = \{(-a,b),(a,b)\}$. Once again, due to the geometric interpretation, such graphs don't intersect when $k < - 5$.

Here it is a bonus question: for which values of $k$ does the graph of $f(x) = x^{2} - 4$ intersects the graph of $g(x) = -2|x| + k$ exactly once and exactly twice, respectively?

For $x$ and $k$ real numbers, determine the values of $k$ for which the graphs of $f(x)=x^2-4$ and $g(x)=2|x|+k$ do not intersect

3 Answers3