Determining the signs of certain expressions formed by the coefficients of a quadratic (Graph of quadratic given)

Question

Consider the quadratic $y=ax^2+bx+c$ with the following graph:

I'm trying to figure out the sign of the following expressions:

1). $\frac{c}{a}$

2). $b+4a$

3). $2a+b$

The vertex form may prove to be useful:

$y=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$

Also, since there are two real x-intercepts, we know that the discriminant is positive.

let $x_r$ denote the lesser root, so that $1<x_r<2$.

Using the quadratic formula to express $x_r$ we see that:

$1<(-b+\sqrt{b^2-4ac})\frac{1}{2a}<2$

(note that since $a<0$, this is indeed the lesser root)

This leads to the inequality:

$2a+b>\sqrt{b^2-4ac}>4a+b$

Thus $2a+b>0$

I'm a little stuck on (1) and (2). Help appreciated! Also, alternative methods to do (3) are definitely appreciated (as well as errors in my argument)

Answer 1

They’re all positive:

1. by Vieta’s formulae
2. by considering the derivative/gradient at $x=2$
3. by considering the derivative/gradient at $x=1.$

Answer 2

Since the parabola is "downward-opening", we have $ \ a \ < \ 0 \ \ . \ $ If we label the two $ \ x-$intercepts as $ \ r \ $ and $ \ s \ \ , \ $ then the graph indicates that $ \ 1 \ < \ r \ < \ 2 \ $ and $ \ 2 \ < \ s \ \ , $ implying $ \ 2 \ < \ s \ < \ r·s \ < \ 2·s \ \ . $ The factorization of the polynomial is then $$ \ a·(x - r)·(x - s) \ \ \Rightarrow \ \ ax^2 \ + \ bx \ + \ c \ \ = \ \ ax^2 \ - \ a(r + s)·x \ + \ a·r·s \ \ . \ $$ As we have found that $ \ r·s \ > \ 0 \ \ , \ $ then $ \ c \ $ and $ \ a \ $ must have the same sign, so $ \ \frac{c}{a} \ > \ 0 \ \ . $ (This argument is somewhat more reliable at this stage than using the positivity of the $ \ y-$coordinate of the vertex to establish that $ \ b^2 - 4ac \ > \ 0 \ \ , \ $ since we know that $ \ 4ac \ > \ 0 \ \ , \ $ but have not determined the relative "magnitude" of $ \ b^2 \ . \ ) $

Applying the vertex form of the parabola, we have $ \ x_{vert} \ = \ -\frac{b}{2a} \ \ $ and we also know that $ \ x_{vert} \ = \ \frac{r \ + \ s}{2} \ \ . \ $ From the above inequalities, we find $ \ 3 \ < \ r + s \ = \ \ 2·x_{vert} \ \ , $ thus $$ -\frac{b}{2a} \ \ = \ \ \frac{b}{2·|a|} \ \ > \ \ \frac32 \ \ \Rightarrow \ \ b \ \ > \ \ 3·|a| \ \ \Rightarrow \ \ b \ - \ 3·|a| \ \ > \ \ 0 \ \ ; $$ since $ \ a \ < \ 0 \ \ , \ $ we can conclude that $ \ b + 2a \ > \ b + 3a \ > \ 0 \ \ . $ We can go a bit farther, since the graph shows $ \ x_{vert} \ > \ 2 \ \ , \ $ hence $$ \frac{b}{2·|a|} \ \ > \ \ 2 \ \ \Rightarrow \ \ b \ - \ 4·|a| \ \ > \ \ 0 \ \ \Rightarrow \ \ b \ + \ 4a \ > \ 0 \ \ . $$

(One does need to be a little careful: because $ \ a \ $ is negative, this last result doesn't automatically follow from the preceding one; rather, we can say $ \ b + 2a \ > \ b + 4a \ > \ 0 \ \ . \ ) $