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Nature of Solutions Graphically

If a > 0 an...

Question

If $a > 0$ and $b^{2} - 4 a c > 0$ , then the graph of $y = a x^{2} + b x + c$

A

lies entirely above the x-axis.

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B

is concave upwards and cuts the x-axis.

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C

touches the x-axis and lies below it.

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D

is concave downwards.

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Solution

The correct option is B is concave upwards and cuts the x-axis.
It has two real roots. So, it cuts x-axis twice.

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a < 0

D < 0

, then the graph of

y = a x^{2} + b x + c

(where D = b^{2} - 4 a c)

a > 0

b^{2} - 4 a c = 0

, then the graph of

y = a x^{2} + b x + c

a < 0

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, then the graph of

y = a x^{2} + b x + c

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then the graph of

y = a x^{2} + b x + c

b^{2} - 4 a c > 0

, then the graph of

y = a x^{2} + b x + c

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