There are answers where they assume that $ y$ and the discriminant are $0$. But why is that? By finding the values through the vertex form, the value of $y$ or the discriminant isn't $0$.
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1Let suppose $y=ax^2+bx+c$ has vertical symmetry axis with some $x=x_0$. Then for any $x_1$ if $(x_1,y_1)$ belongs to plot then $(2x_0-x_1,y_1)$ must belong to plot. $$y_1=ax_1^2+bx_1+c \Rightarrow y_1=a(2x_0-x_1)^2+b(2x_0-x_1)+c$$ $$ax_1^2+bx_1+c=a(2x_0-x_1)^2+b(2x_0-x_1)+c \forall x_1$$ $$4ax_0^2-4ax_0x_1+2bx_0-2bx_1 = 0 \forall x_1$$ $$4ax_0^2+2bx_0=0, 4ax_0+2b=0$$ $$x_0=-\frac{b}{2a}$$ – Ivan Kaznacheyeu Jun 08 '22 at 13:52
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We are not looking for solutions to the equation $y = ax^2 +bx + c$. We are looking for the lowest point of the graph.
The parabola is symmetric about the vertical line through the lowest point. Differentiate to see the minimum has $x$-coordinate $-b/2a$.
If you don't know calculus you can instead complete the square to get $$ ax^2 +bx + c = a(x +b/2a)^2+ \text{some constant}$$
Clearly the above is minimised at $x=-b/2a$.
Daron
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@PierreCarre Yep. On the other hand for $a<0$ we are looking for the maximum and the reasoning is the same. – Daron Jun 08 '22 at 14:29
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The line $x=-\frac{b}{2a}$ is a line of symmetry because, if we take $f(x)=ax^2+bx+c$, then, $$f(-\frac{b}{2a}+\epsilon)=a\left(-\frac{b}{2a}+\epsilon\right)^2+b\left(-\frac{b}{2a}+\epsilon\right)+c$$ $$=-\frac{b^2}{4a}+a\epsilon^2+c= f(-\frac{b}{2a}-\epsilon) $$ Therefore the value of f(x) for any $x_1$, $x_2$ such that $x_1$, $x_2$ are the same distance from $-\frac{b}{2a}$ but on opposite sides, is the same.
insipidintegrator
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David Quinn
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I appreciate your answer, but it is beyond my intellect. I am a high school student. – user324713 Jun 09 '22 at 04:53