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I was doing quadratic equations and their graphical representation where I was told that if a>0, the parabola faces in a certain direction whereas for a<0, the parabola faces a different direction. Now, how does the value of 'a' affect the direction of the parabola ?

Rayhan
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  • You should at least tell us what $a$ is here. I suspect coëfficient of $x^2$. If so then make pictures of $y=x^2$ and $y=-x^2$ and compare. – drhab Aug 16 '22 at 15:30
  • When $x$ is large enough positive or negative number, $|ax^2| \gg |bx| \gg |c|$. That's why sign of expression $ax^2+bx+c$ is determined by sign of $ax^2$ which is the same as sign of $a$. When $a$ is positive $ax^2$ is positive, then branches of parabola are directed to positive direction of $y$-axis. When $a$ is negative $ax^2$ is negative then branches of parabola are directed to negative direction of $y$-axis. – Ivan Kaznacheyeu Aug 17 '22 at 11:52

2 Answers2

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First sketch $y=a*x^2$ for a chosen positive a, then for a negativ a. then move the graph up or down bei b so you have $ y=a*x^2+b $ than move it to right or left by substituting x with (x-c) and you have $y=a*(x-c)^2+b$ since you only moved your graph it always looks in the same direction.

trula
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Consider $y = ax^2+bx+c, \quad a < 0$. Then, this is the same as $$y = -|a|x^2+bx+c = -(\sqrt{|a|}x-\frac{b}{2\sqrt{|a|}})^2+c+\frac{b^2}{4|a|}.$$ Because the first term on the right hand side is the negative of a square, the maximum value of $y$ is $c+\frac{b^2}{4|a|}$, and any perturbation of away from $x = b/2|a|$ will make the value of $y$ decrease. Hence, the parabola must face down in this case.

You can do similar analysis with $a>0$ to show that parabola would face upwards.

Doug
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