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I need some help figuring out how to solve this problem.

Which of the following could be a graph of the equation $ y= ax^2 + bx + c$ where $b^2 - 4ac = 0$

The picture below was the correct answer. So the first equation told me that the the shape of the graph would be a parabola. Then the second equation is what I assume tells you that this parabola is shifted to the right and is facing upwards although I don't see what about $b^2 - 4ac = 0$ tells me this. Answer

j76goatboy
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1 Answers1

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I'm assuming that it's $ax^2+bx+c$, not $ax^2+bx+x$.

Actually, nothing you've included tells you that it's shifted to the right or facing upward. It faces upward if and only if $a > 0$. It's shifted to the right if and only if $b < 0$ (given that $a > 0$). What $b^2-4ac = 0$ tells you is that it's tangent to the $x$-axis. (Equivalently, it has a double root.)

I'm assuming this was a multiple-choice problem. The critical part would have been recognizing what $b^2-4ac = 0$ tells you, and to identify the one parabola that was tangent to the $x$-axis.

Brian Tung
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  • Why does that tell you that its tangent to the x axis? Does $b^2 - 4ac =$ the y position of the vertex? – j76goatboy Dec 09 '15 at 01:06
  • No. Are you familiar with the quadratic formula? The two solutions to $ax^2+bx+c = 0$ are $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-fac}}{2a}$. If $b^2-4ac = 0$, then the square root is equal to $0$, and the two solutions are identical; i.e., the parabola is tangent to the $x$-axis. – Brian Tung Dec 09 '15 at 01:11
  • The $y$ position of the vertex is the $y$-value corresponding to $x = -\frac{b}{2a}$. – Brian Tung Dec 09 '15 at 01:12