Question

The graphs of g(x) = x³- ax² + 6 and h(x) = 2x² + bx + 3 touch when x = 1. Therefore, the tangent to
the curve of g at x = 1 is also the tangent to the curve of h at x = 1.
Determine the coordinates of this point of contact of the two graphs.​

Answer 1

Answer:

$(1,\, 10)$.

Step-by-step explanation:

Differentiate each function to find an expression for its gradient (slope of the tangent line) with respect to $x$. Make use of the power rule to find the following:

$g^\prime(x) = 3\, x^2 - 2\, a\, x$.

$h^\prime(x) = 2\, (2\, x) + b = 4\, x + b$.

The question states that the graphs of $g(x)$ and $h(x)$ touch at $x = 1$, such that $g^\prime(1) = h^\prime(1)$. Therefore:

$3 - 2\, a = 4 + b$.

On the other hand, since the graph of $g(x)$ and $h(x)$ coincide at $x = 1$, $g(1) = h(1)$ (otherwise, the two graphs would not even touch at that point.) Therefore:

$1 - a + 6 = 2 + b + 3$.

Solve this system of two equations for $a$ and $b$:

$\begin{aligned}& a + b = 2 \\ & 2\, a + b = -1\end{aligned}$.

Therefore, $a = -3$ whereas $b = 5$.

Substitute these two values back into the expression for $g(x)$ and $h(x)$:

$g(x) = x^3 + 3\, x^2 + 6$.

$h(x) = 2\, x^2 + 5\, x + 3$.

Evaluate either expression at $x = 1$ to find the $y$-coordinate of the intersection. For example, $g(1) = 1 + 3 + 6 = 10$. Similarly, $h(1) = 2 + 5 + 3 = 10$.

Therefore, the intersection of these two graphs would be at $(1,\, 10)$.