TITLE: Spivak, Ch. 4 Graphs, Problem 15: Draw the graph of $f(x)=ax^2+bx+c$?
QUESTION [2 upvotes]: I am asking about this problem because the solution manual seems to have a much more limited solution than mine below, and I wonder if they have been lazy or if I have done something incorrect.

Draw the graph of $f(x)=ax^2+bx+c$.Hint: use methods of Problem 1-18.

We would like to draw the graph, but since we are only in chapter 4 of Spivak's Calculus, we have limited methods available to draw conclusions about $f$.
Here is the solution from the solution manual:
$$f(x)=ax^2+bx+c = a\left(x^2+\frac{b}{a}x+\frac{c}{a} \right)$$
$$=a\left[ \left(x+\frac{b}{2a} \right)^2 + \left(\frac{c}{a}-\frac{b^2}{4a} \right) \right]$$
the graph looks like

Here is my solution
Factorize $f(x)$
$$f(x)=a(x^2+\frac{b}{a}x+\frac{c}{a})$$
What do we know about $g(x)=x^2+\frac{b}{a}x+\frac{c}{a}$? Let's factorize it
$$g(x)=x^2+2x\frac{b}{2a}+\left( \frac{b}{2a} \right)^2+c-\left( \frac{b}{2a} \right)^2$$
$$=\left(x+\frac{b}{2a}\right)^2 + c -\left( \frac{b}{2a} \right)^2$$
$$=\left(x+\frac{b}{2a}\right)^2+ \frac{4ac-b^2}{4a^2}\tag{1}$$
Note that $g(x)$ has a minimum value when the term $\left(x+\frac{b}{2a}\right)^2$ is zero, ie when $x=\frac{-b}{2a}$.
Set to zero to find roots
$$g(x)=0 \implies \frac{b^2-4ac}{4a^2}=\left(x+\frac{b}{2a}\right)^2$$
The right side is nonnegative. For there to be a solution (ie for $g(x)$ to have a root), the left side must also be nonnegative.
Case 1: The left side is zero, ie $b^2=4ac$ $\implies$ There is only one root
$$x_r = \frac{-b}{2a}$$
The minimum of $g(x)$ is at $x_r$.
Case 2: $b^2-4ac>0 \implies$ There are two roots
$$x_r=\frac{-b \pm\sqrt{b^2-4ac}}{2a}$$
Case 3: $b^2-4ac<0 \implies$ There are no roots, $g(x)>0\ \forall x$
$$g(x) = \left(x+\frac{b}{2a}\right)^2+ \frac{4ac-b^2}{4a^2}>0,\  \forall x$$
What do we know about $f(x)=a(x^2+\frac{b}{a}x+\frac{c}{a})$ now?
Case 1: $a>0$, $b^2-4ac=0 \implies$ one root at the minimum, $f(x)$ grows without bound as $x \to \pm \infty$.
Case 2: $a<0$, $b^2-4ac=0 \implies$ case 1 multiplied by a negative number. One root at the maximum, $f(x)$ decreases without bound as $x \to \pm \infty$.
Case 3: $a>0$, $b^2-4ac>0 \implies$ similar to case 1, but with two roots
Case 4: $a<0$, $b^2-4ac>0 \implies$ similar to case 2 but with two roots
Case 5: $a>0$, $b^2-4ac<0 \implies$ similar to case 1 but with no roots
Case 6: $a<0$, $b^2-4ac<0 \implies$ similar to case 2 but with no roots

REPLY [3 votes]: You're both correct. Spivak's diagram is perfectly fine, as long as you accept the possibility that:

The axes are not necessarily drawn over the lines $x = 0$ and $y = 0$; and

The axes are not necessarily increasing in the usual directions.


These are both generally considered bad things to do in most cases because of the misleading implications, but they're definitely things you can do.
Notice that your various cases are translations, flips and scalings of the shape given in the answer, which are all effects you can also achieve by redefining the axes appropriately.
Oh, and you're both assuming that $a \neq 0$, otherwise the graph is just a line.
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