Quadratics, transformations, and formulas

Question

Two-part question. Feel free to answer just one part, or both (write which letter part you are answering)

a) If the quadratic function $g(x)=a(x-h)^2+ k$ does not touch the $x$-axis, what can be said about $a$ and $k$? There are two cases.

I'm having trouble explaining the cases. Please use transformations to explain the answer

So far, I only know the basics of $k$ shifting the graph up, so if $k$ does not touch the $x$-axis, it is any real number except $0$. I also know $a$ can be any real number, making the graph taller or wider. What else can be said about $a$ and $k$?

b) If the quadratic function $g(x)=a(x-h)^2+ k$ touches the $x$-axis exactly once, what can you say about $a$ and $k$?

Answer 1

The graph is a parabola. If a>0, then the parabola goes up and k has to be k>0 so that the vertex is above the x-axis. (The vertex is (h,k) and k moves the vertex up in this case) If a<0, then the parabola goes down and k has to be k<0 so that the vertex is below the x-axis. (k moves the vertex down in this case--because k is neg. Those are the two cases: when a>0 and when a<0.

Answer 2

Your analysis is correct. The same reasoning applies to both the problems.To know more about the relation between a and k you will have to use algebra. For the first case the discriminant has to be less than 0 and for the second the discriminant has to be 0.

$$g(x)=a(x-h)^2+k$$ $$\Rightarrow g(x)=ax^2-2axh+k+ah^2$$

If g(x) never touches the x-axis, it means there exists no value of x for which g(x)=0 and if g(x) touches the x-axis it would imply that there is only one value of x for which g(x)=0.For discriminant click me. Can you take it from here?

Answer 3

Let me add to GTX OC's answer. Please give him/her the credit. Also, upvote any answer that helps you and accept any that solves the problem. I think GTX gave the answer but here is some more explanation.

From what GTX wrote

$$ g(x)=ax^2-2axh+k+ah^2$$

The discriminant is $$ \Delta = (2 a h)^2 - 4 a (k+a h^2) = -4 a k$$

Now we have three cases $$ \Delta ~~\left\{ \begin{array}{ll} < 0, & \hbox{$g(x)$ does not touch the $x$ axis;} \\ =0, & \hbox{$g(x)$ touches $x$ axis exactly at one point;} \\ >0, & \hbox{$g(x)$ crosses $x$ axis at two points.} \end{array} \right.$$

Now $a \neq 0$ for it to be a quadratic. So what can you say about $k$ in each of the cases?

I realize you are new here. All of us do this because we want to help. So do the right thing and vote up any answer that helps you and accept any that you feel is the best for your needs.

Answer 4

(a) Personally, I prefer the simplest solution, which often highlights an important fundamental concept. We're dealing with an inequality here, so a handy point to notice in this case is that $(x-h)^2\ge 0$ for all $x \in \mathbb R$.

Note: I have edited this answer to include some references to the topic of transformations as this aspect has been highlighted in the OP. Hopefully this hasn't obscured my main point as mentioned above.

Hence, either (i) $a(x-h)^2 \ge 0$ for all $x \in \mathbb R$ if $a>0$ or (ii) $a(x-h)^2 \le 0$ for all $x \in \mathbb R$ if $a<0$.

Now, $a(x-h)^2 = 0$ when $x=h$, so there's a point when the graph would touch the x-axis, if $k = 0$.

In terms of transformations, if $y = f(x)$ in general, then $y = f(x) + k$ is a vertical translation in the direction of the sign of $k$. If $k>0$, it translates upwards, if $k<0$ it translates downwards.

So, in case (i), $k>0$ would mean an upward translation and $a(x-h)^2 + k > 0$ for all $x \in \mathbb R$. In case (ii), $k<0$ would mean a downward translation and $a(x-h)^2 + k < 0$ for all $x \in \mathbb R$.

As in each case, with $k=0$ there would be one point on the graph touching the $x$-axis, the translation takes the graph in the appropriate direction away from the $x$-axis so as not to intersect.

With $y=af(x)$, we have an enlargement if $a>0$. If $a<0$ it is an enlargement combined with a reflection in the $x$-axis. An important point in this problem is that the enlargement doesn't change the basic solution because if $f(x)=0$ then $af(x)=0$ and the reflection just gives us the alternative case.

The answer to (b) is contained within my answer to (a).

Note: There is of course the trivial case of a=0, but I'll leave you to ponder that.