The diagram shows the graph of y - ax 2- bx - c- Then A- a -0B- b -0C- b2-4 a c-0D- c -0

The correct option is B b lt; 0∵ a 0 Hence x_2+x_1 0 (from figure) ⇒ a/b 0 ⇒a/b⇒ b 0 (∵ a 0). The curve y=ax^2+bx+c meets y axis at x=0. Therefore y=c From ...

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Standard XII

Mathematics

Graph of Quadratic Expression

The diagram s...

Question

The diagram shows the graph of $y = a x^{2} + b x + c$ . Then

a > 0

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b < 0

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c < 0

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b² - 4ac = 0

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Solution

The correct option is B b < 0
$∵ a < 0$
Hence $x_{2} + x_{1} < 0$ (from figure)
$\Rightarrow \frac{- a}{b} < 0 \Rightarrow \frac{a}{b} \Rightarrow b < 0 (∵ a < 0)$ .
The curve $y = a x^{2} + b x + c$ meets $y - a x i s$ at $x = 0$ .
Therefore $y = c$ From the figure $y > 0 \Rightarrow c > 0$

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Q. If the diagram in Fig. 2.22 shows the graph of the polynomial f(x) = ax² + bx + c, then

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The diagram shows the graph of y=ax2+bx+c. Then

The correct option is B b < 0∵ a<0 Hence x2+x1<0 (from figure) ⇒−ab<0⇒ab⇒b<0(∵a<0). The curve y=ax2+bx+c meets y−axis at x=0. Therefore y=c From the figure y>0⇒c>0

The diagram shows the graph of $y = a x^{2} + b x + c$ . Then

The correct option is B b < 0
$∵ a < 0$
Hence $x_{2} + x_{1} < 0$ (from figure)
$\Rightarrow \frac{- a}{b} < 0 \Rightarrow \frac{a}{b} \Rightarrow b < 0 (∵ a < 0)$ .
The curve $y = a x^{2} + b x + c$ meets $y - a x i s$ at $x = 0$ .
Therefore $y = c$ From the figure $y > 0 \Rightarrow c > 0$