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The equation of a parabola is given by: $y= ax^2 + bx+c$

Why is it that when the coefficient of $x$ i.e. $a$ is positive we get an upward parabola and when it's negative we get a downward parabola?

Also, I saw that increasing the value of $|a|$ narrows the parabola, why?

Lastly, what is the role of $b$ in determining the structure of this parabola?

Archer
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3 Answers3

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In the quadratic equation "$b$" and "$c$" terms are correlated to an axis translation, thus we can consider the simpler case $$y=ax^2$$ for which is clear the role of "$a$" to determine the sign of $y$.

To clarify the first point suppose to change the coordinates by translation by means of $y=(y+k)$ and $x=(x+h)$ then $$(y+k)=a(x+h)^2$$ $$y=ax^2+2hx+h^2-k$$ which is in the form $$y=ax^2+bx+c$$

I think this way is simpler because you don't need any calculus knowledge.

user
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  • Elegant answer but please explain the role of $b$.I know that c determines the "intercept" of the parabola. – Archer Nov 27 '17 at 10:02
  • starting from the simplest case $y=ax^2$ if you translate x by h you obtain $y=ax^2+2hx+h^2$ thus h=b/2 is the translation of the vertex along the x axis (on the right if h is negative) – user Nov 27 '17 at 10:07
  • Why increasing the value of $a$ narrows the parabola? Can you please answer that too :) , that was also a part of my question – Archer Mar 15 '18 at 21:19
  • @Abcd in the equation $y=ax^2$ fix two values, for example x=-1 and x=1 to which correspond y=a. Now if you double "a\to 2a" we obtain $y=2a$. Can you see why is is narrow? – user Mar 15 '18 at 21:29
  • yes, thanks :) . – Archer Mar 15 '18 at 22:10
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    @Abcd It was one of my first answer here on MSE! You are welcome! Bye – user Mar 15 '18 at 22:11
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let $$f(x)=ax^2+bx+c$$ then we have $$f'(x)=2ax+b$$ and $$f''(x)=2a$$ if $$a>0$$ then we get a Minimum Point, if $a<0$ then we get a Maximum Point.

Dr. Sonnhard Graubner
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You're given $y = a x^2 + b x + c $

Start your analysis by completing the square in $x$, as follows:

$\begin{equation} \begin{split} y &= a ( x^2 + \dfrac{b}{a} x ) + c \\ &= a (x + \dfrac{b}{2a} )^2 - a (\dfrac{b}{2a})^2 + c \\ &= a (x - x_0)^2 + y_0 \\ \end{split} \end{equation}$

So that,

$ (y - y_0) = a (x - x_0)^2 $

If $a \gt 0 $, then as $x $ gets away from $x_0$ in either direction (left or right), $y$ will increase above $y_0$, and if we increase $a$ then $y$ will grow faster, thus for the same value of $|x - x_0|$ , $y $ will be higher, thus making the graph look narrower. The same can be said when $a \lt 0$ where deviations of $x$ from $x_0$ cause $y$ to dip below $y_0$ , and as we make the negative $a$ more negative, the deviation of $y$ from $y_0$ will increase, again making the graph narrower.

Hosam Hajeer
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