Quadratic Formula: If a is negative and c is positive, what can be said about the graph of y = ax^2 + bx + c?

[ConnorK]

Sorry about the blurry picture. The answer i got is correct but I'm not sure if i got there correctly.

 

[blamocur]

If I were grading your work I'd have two issues:

1. First you say $\Delta = b^2 - 4ac$, then $\Delta = b^2+4ac$, which makes no sense to me.
2. You are right that $\Delta > 0$, but you can add to the description of the graph by comparing values of $y$ for $x=0$ and for large values of $x$.

 

[Steven G]

To continue with what @blamocur stated, you should say that b²>0 and -4ac>0 so that b² -4ac>0

 

[fresh_42]

Two things can always be done with $y=ax^2+bx+c$. One is the completion of the square:
$$\begin{array}{lll} y&=ax^2+bx+c=a\left(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\right)=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a} \end{array}$$which already gives you the information on whether the parabola is open at the top or bottom and whether it has zeroes.
It is also why we consider $\Delta = b^2 -4ac$. It decides whether $y=0$ has none, one or two solutions.

The other is differentiation. $y'(x)=2ax+b$ and $y''(x)=2a.$ The solution of the equation $y'(x)=0 \Longleftrightarrow x=-\dfrac{b}{2a}$ tells us where the parabola has its maximum or minimum, depending on whether $y''(x)=2a$ is positive or negative. In your case, it is assumed to be negative, so the parabola is open at the bottom.

Mnemonic: $y=x^2$ opens at the top, so all negative parabolas $y=-x^2+\ldots$ open at the bottom.

 

[Aion]

Two theorems come to my mind regarding this topic that don't invoke calculus.

120. For all real values of $x$, the expression $ax^2+bx+c$ has the same sign as $a$, except when the roots of the equation $ax^2+bx+c=0$ are real and unequal, and $x$ has a value lying between them.

121. From the preceding article, it follows that the expression $ax^2+bx+c$ will always have the same sign whatever real value x may have, provided $b^2-4ac$ is negative or zero; and if this condition is satisfied the expression is positive or negative according as $a$ is positive or negative. Conversely, so that the expression $ax^2+bx+c$ may be always positive, $b^2-4ac$ must be negative or zero, and $a$ must be positive; and in order that $ax^2+bx+c$may be always negative $b^2-4ac$ must be negative or zero, and $a$ must be negative.

p.90 Higher Algebra by Hall & Knight.​

 

[Aion]

Two theorems come to my mind regarding this topic that don't invoke calculus.

120. For all real values of $x$, the expression $ax^2+bx+c$ has the same sign as $a$, except when the roots of the equation $ax^2+bx+c=0$ are real and unequal, and $x$ has a value lying between them.

121. From the preceding article, it follows that the expression $ax^2+bx+c$ will always have the same sign whatever real value x may have, provided $b^2-4ac$ is negative or zero; and if this condition is satisfied the expression is positive or negative according as $a$ is positive or negative. Conversely, so that the expression $ax^2+bx+c$ may be always positive, $b^2-4ac$ must be negative or zero, and $a$ must be positive; and in order that $ax^2+bx+c$may be always negative $b^2-4ac$ must be negative or zero, and $a$ must be negative.

p.90 Higher Algebra by Hall & Knight.​

It is common to refer to the graph of a quadratic function in the case of $a>0$ as an up parabola, and in the case of $a<0$ as a down parabola. However, that assumes that the definition of a parabola is known in the first place. First, one must prove that the graph of a quadratic function is a parabola.

 

[Harry_the_cat]

Yes you can show it algebraically as you have attempted. Steven G has shown you how to fix it up, so each statement is correct.

Alternatively, you could think about it graphically:

If a<0, then the graph is an "upside-down" parabola.

If c>0, then, seeing that c is the y-intercept, the parabola cuts the y-axis above the origin.

Now, stop and think about those 2 things. Can you see that the graph MUST have 2 x-intercepts? (Try to draw it otherwise.)

Note also: You concluded "there are 2 solutions to the graph in the question".

Graphs don't have solutions. Equations do though.

You should have concluded:

There are 2 solutions to the equation $\displaystyle ax^2+bx+c=0$ which means that the graph of $\displaystyle y=ax^2+bx+c$ has 2 x-intercepts.