Suppose a parabola \[y=a{{x}^{2}}+bx+c\] has two x intercepts one positive and one negative, and its vertex is (2,-2). Then which of the following is true?
(a) ab>0
(b) bc>0
(c) ca>0
(d) a+b+c>0
Answer
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Hint: Vertex is at (2, -2) and the parabola cuts the x axis so it is an upward opening parabola. So we will use this information to draw the figure. And as the slope is zero we will get the relation between a and b by differentiating.
Complete step-by-step answer:
So it is mentioned in the question that the vertex is at (2, -2) and the parabola cuts the x axis once positive axis and once negative axis so it is an upward opening parabola.
So according to the details provided in the question the figure is:
And we know that for an upward opening parabola a>0.
And at the vertex (2, -2) the slope is zero which means \[\dfrac{dy}{dx}=0\]. Hence using this information, we get,
\[y=a{{x}^{2}}+bx+c.......(1)\]
So differentiating equation (1) with respect to x we get,
\[\dfrac{dy}{dx}=2ax+b.......(2)\]
Now equating equation (2) with zero we get.,
\[2ax+b=0.......(3)\]
Now substituting 2 in place of x and solving for b we get,
\[\begin{align}
& \Rightarrow 2a\times 2+b=0 \\
& \Rightarrow b=-4a.........(4) \\
\end{align}\]
Now we know that a>0 for upward parabola then from equation (4) we can clearly see that b<0. And for such a parabola we can clearly see from the figure that the y intercept is positive.
So here a>0, c<0 and from equation (4) b<0.
So ab<0 as we are multiplying a positive quantity with negative quantity.
So bc>0 as we are multiplying a negative quantity with negative quantity.
And ca<0 as we are multiplying a positive quantity with negative quantity.
And a+b+c<0 as both b and c are negative.
So the right answer is option (b).
Note: Drawing the figure is important as it makes everything clear. Also we need to remember that slope means differentiation of y with respect to x. We in a hurry can make a mistake in understanding the relation between a, b and c so we need to be careful while doing this step.
Complete step-by-step answer:
So it is mentioned in the question that the vertex is at (2, -2) and the parabola cuts the x axis once positive axis and once negative axis so it is an upward opening parabola.
So according to the details provided in the question the figure is:
And we know that for an upward opening parabola a>0.
And at the vertex (2, -2) the slope is zero which means \[\dfrac{dy}{dx}=0\]. Hence using this information, we get,
\[y=a{{x}^{2}}+bx+c.......(1)\]
So differentiating equation (1) with respect to x we get,
\[\dfrac{dy}{dx}=2ax+b.......(2)\]
Now equating equation (2) with zero we get.,
\[2ax+b=0.......(3)\]
Now substituting 2 in place of x and solving for b we get,
\[\begin{align}
& \Rightarrow 2a\times 2+b=0 \\
& \Rightarrow b=-4a.........(4) \\
\end{align}\]
Now we know that a>0 for upward parabola then from equation (4) we can clearly see that b<0. And for such a parabola we can clearly see from the figure that the y intercept is positive.
So here a>0, c<0 and from equation (4) b<0.
So ab<0 as we are multiplying a positive quantity with negative quantity.
So bc>0 as we are multiplying a negative quantity with negative quantity.
And ca<0 as we are multiplying a positive quantity with negative quantity.
And a+b+c<0 as both b and c are negative.
So the right answer is option (b).
Note: Drawing the figure is important as it makes everything clear. Also we need to remember that slope means differentiation of y with respect to x. We in a hurry can make a mistake in understanding the relation between a, b and c so we need to be careful while doing this step.
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