If $(2, 6)$ $(2,6)$ lies on the curve $f (x) = a x^{2} + b x$ $f(x) = ax^2+bx$ and $y = x + 4$ $y=x+4$ is a tangent to the curve at that point. Find $a$ $a$ and $b$ $b$ ?

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Answer 1 · 2017-12-31T21:22:37

$a = - 1, b = 5$ $a=-1, b=5$

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

We have:

$f (x) = a x^{2} + b x$ $f(x) = ax^2+bx$

We are given that the point $(2, 6)$ $(2,6)$ lies on the curve:

$\Rightarrow f (2) = 6$ $=> f(2)=6$
$:. 4a+2b = 6$
$:. 2a+b = 3$ ..... [A]

If we differentiate the parabola equation, then we have:

$f'(x) = 2ax+b$

The gradient of the tangent at $(2,6)$ is given by:

$m_T = f'(2) = 4a+b$

We also know that the tangent equation is $y=x+4$ and so comparing with the standard straight line equation $y=mx+c$ :

$=> m_T = 1$
$:. 4a+b = 1$ ..... [B]

We now solve the equations [A] and [B] simultaneous:

$[B]-[A] => 2a = -2 => a=-1$

Substitute $a=-1$ into [A]:

$b-2=3 => b=5$

Hence we have:

$a=-1, b=5$

If (2,6)(2,6)(2,6) lies on the curve f(x) = ax^2+bx f(x)=ax2+bxf(x) = ax^2+bx and y=x+4y=x+4y=x+4 is a tangent to the curve at that point. Find aaa and bbb?