If the curve y=ax^2-6x+b passes through (0,4) | vedclass

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

Question

(A) Given the curve equation is $y = ax^2 - 6x + b$.Since the curve passes through $(0, 4)$,we substitute $x=0$ and $y=4$ into the equation:$4 = a(0)^

Vedclass · Accepted Answer

$2, 4$

Vedclass · Answer

(A) Given the curve equation is $y = ax^2 - 6x + b$.Since the curve passes through $(0, 4)$,we substitute $x=0$ and $y=4$ into the equation:$4 = a(0)^2 - 6(0) + b \implies b = 4$.Now,the derivative of the curve is $\frac{dy}{dx} = 2ax - 6$.The tangent is parallel to the $x$-axis at $x = \frac{3}{2}$,which means $\frac{dy}{dx} = 0$ at $x = \frac{3}{2}$.$2a(\frac{3}{2}) - 6 = 0 \implies 3a - 6 = 0 \implies 3a = 6 \implies a = 2$.Thus,the values are $a = 2$ and $b = 4$.

If the curve $y=ax^2-6x+b$ passes through $(0,4)$ and has its tangent parallel to the $x$-axis at $x=\frac{3}{2}$,then the values of $a$ and $b$ respectively are

Similar Questions

The length of the subtangent at $(2, 2)$ to the curve $x^5 = 2y^4$ is

The equation of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at the point $(x_1, y_1)$ is:

What are the lengths of the intercepts on the coordinate axes made by the tangent to the curve $y = 2x^2 + 3x - 2$ at the point $(1, 3)$?

The subnormal at any point on the curve $xy^n = a^{n + 1}$ is constant for $n=$ ............

If the tangent drawn at a point $P$ on the curve $y=3x^2-5x+7$ is parallel to its chord joining the points $(1, y_1)$ and $(2, y_2)$ on it,then the $x$-coordinate of the point $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module