The curve y=ax^3+bx^2+cx+5 touches the X-axis | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $X$-axis at $P(-2,0)$,the point $(-2,0)$ lies on the curve and the gradient at this point

Vedclass · Accepted Answer

$a=\frac{-1}{2}, b=\frac{-3}{4}, c=3$

Vedclass · Answer

(D) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $X$-axis at $P(-2,0)$,the point $(-2,0)$ lies on the curve and the gradient at this point is $0$. Substituting $P(-2,0)$ into the equation: $0 = a(-8) + b(4) + c(-2) + 5 \Rightarrow -8a + 4b - 2c = -5 \Rightarrow 8a - 4b + 2c = 5 \quad (1)$. The derivative is $\frac{dy}{dx} = 3ax^2 + 2bx + c$. At $P(-2,0)$,$\frac{dy}{dx} = 0$: $3a(-2)^2 + 2b(-2) + c = 0 \Rightarrow 12a - 4b + c = 0 \quad (2)$. The curve cuts the $Y$-axis at $Q(0,k)$. At $Q$,$x=0$ and the gradient is $3$: $\frac{dy}{dx}|_{x=0} = 3(a)(0)^2 + 2(b)(0) + c = 3 \Rightarrow c = 3$. Substituting $c=3$ into $(1)$ and $(2)$: $(1): 8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1 \quad (3)$. $(2): 12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3 \quad (4)$. Subtracting $(3)$ from $(4)$: $(12a - 8a) = -3 - (-1) \Rightarrow 4a = -2 \Rightarrow a = -\frac{1}{2}$. Substituting $a = -\frac{1}{2}$ into $(3)$: $8(-\frac{1}{2}) - 4b = -1 \Rightarrow -4 - 4b = -1 \Rightarrow -4b = 3 \Rightarrow b = -\frac{3}{4}$. Thus,$a = -\frac{1}{2}, b = -\frac{3}{4}, c = 3$.

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $P(-2,0)$ and cuts the $Y$-axis at a point $Q$,where its gradient is $3$. Then:

Similar Questions

The sum of the squares of the intercepts on the axes made by a tangent at any point on the curve $x^{2/3} + y^{2/3} = a^{2/3}$ is -

What is the equation of the tangent to the curve $y = 2 \cos x$ at $x = \pi / 4$?

Find the equations of the tangent and normal to the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at the point $(0,5)$.

Find the equation of the tangent to the curve $y + \frac{2}{x - 3} = 0$ which has a slope of $2$.

The tangent to the curve $y = e^{2x}$ at the point $(0, 1)$ meets the $x$-axis at

Vedclass Test Series

Exam Paper Generator

Online Exam Module