Question

Find the nature of a,b and cof a quadratic polynomials ax2 + bx + c by this graph. ​

Answer 1

There is no graph in this question

Answer 2

SOLUTION

TO DETERMINE

The nature of a , b and c of a quadratic polynomials ax² + bx + c by this graph

EVALUATION

The given polynomial is

$\sf{ y = a {x}^{2} + bx + c\: }$

It represents a parabola

Since the parabola is downwards

So a < 0

The given equation is

$\sf{ y = a {x}^{2} + bx + c\: }$

It cuts y axis

Thus we have x = 0

From above we get y = c

Thus the parabola cuts y axis at (0,c)

Since the point lies on positive y axis

c > 0

Next The given polynomial is

$\sf{ y = a {x}^{2} + bx + c\: }$

It represents a parabola

We rewrite the above polynomial as below

$\displaystyle \sf{ \frac{y}{a} = {x}^{2} + \frac{b}{a} x + \frac{c}{a} \: }$

$\implies \displaystyle \sf{ \frac{y}{a} = { \bigg(x + \frac{b}{2a} \bigg )}^{2} + \frac{c}{a} - \frac{ {b}^{2} }{4 {a}^{2} } \: }$

$\implies \displaystyle \sf{ \frac{y}{a} + \frac{ {b}^{2} }{4 {a}^{2} } = { \bigg(x + \frac{b}{2a} \bigg )}^{2} + \frac{c}{a} \: }$

$\implies \displaystyle \sf{ \frac{y}{a} + \frac{ {b}^{2} }{4 {a}^{2} } - \frac{c}{a} = { \bigg(x + \frac{b}{2a} \bigg )}^{2} \: }$

$\implies \displaystyle \sf{ \frac{1}{a} \bigg( {y}+ \frac{ {b}^{2} }{4 {a}^{} } - c \bigg) = { \bigg(x + \frac{b}{2a} \bigg )}^{2} \: }$

Thus the vertex is

$\displaystyle \sf{ { \bigg( - \frac{b}{2a}, \frac{ {b}^{2} - 4ac}{4a } \bigg )} \: }$

Since the vertex lies on first quadrant

We get

Abscissa > 0

$\displaystyle \sf{ \implies \: - \frac{b}{2a} > 0 }$

$\displaystyle \sf{ \implies \: \frac{b}{2a} < 0 }$

$\displaystyle \sf{ \implies \: b > 0 \: \: \: (\because \: a < 0) }$

FINAL ANSWER

a < 0 , b > 0 , c > 0

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