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miv72 [106K]
3 years ago
9

The graphs of g(x) = x³- ax² + 6 and h(x) = 2x² + bx + 3 touch when x = 1. Therefore, the tangent to

Mathematics
1 answer:
Olegator [25]3 years ago
6 0

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).

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