Question

given the point(2,3) on a tangent line to the curve f(x) = x^2+3x-4 find the points where the tangent touches f(x)




Thing is 2,3 is not on f(x)

Answer 1

Answer:

$\left(2+\sqrt{3},9+7\sqrt{3}\right)\; \textsf{and}\;\left(2-\sqrt{3},9-7\sqrt{3}\right)$

Step-by-step explanation:

Point (2, 3) is on a tangent line to the curve f(x) = x² + 3x - 4.

The slope of the tangent line at any point (x, y) on the curve of f(x) is given by the derivative of f(x) evaluated at x.

Differentiate f(x):

$f'(x)=2x+3$

So, the slope (m) of the tangent line at any point on the curve f(x) can be found by substituting the x-coordinate of the point into m = 2x + 3.

To find the x-coordinates of the points of tangency, where the tangent line at that point passes through (2, 3), substitute the slope of the tangent line (2x + 3) and the point (2, 3) into the point-slope form of a linear equation:

$y-y_1=m(x-x_1)\\\\y-3=(2x+3)(x-2)\\\\y-3=2x^2-x-6\\\\y=2x^2-x-3$

Now, set this equation equal to function f(x) to find the x-coordinates of the points of tangency:

$2x^2-x-3=x^2+3x-4\\\\x^2-4x+1=0$

To solve the quadratic equation, use the quadratic formula:

$\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}$

In this case a = 1, b = -4 and c = 1. Therefore:

$x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(1)}}{2(1)}\\\\\\x=\dfrac{4 \pm \sqrt{16-4}}{2}\\\\\\x=\dfrac{4 \pm \sqrt{12}}{2}\\\\\\x=\dfrac{4 \pm 2\sqrt{3}}{2}\\\\\\x=2\pm \sqrt{3}$

So, the x-coordinates of the points of tangency where the tangent line at those points also passes through point (2, 3) are:

$x = 2+\sqrt{3}\\\\x= 2-\sqrt{3}$

Substitute the found x-values into function f(x) to find the corresponding y-coordinates:

$f(2+\sqrt{3}) = (2+\sqrt{3})^2+3(2+\sqrt{3})-4\\\\f(2+\sqrt{3}) = 7+4\sqrt{3}+6+3\sqrt{3}-4\\\\f(2+\sqrt{3}) = 9+7\sqrt{3}$

$f(2-\sqrt{3}) = (2-\sqrt{3})^2+3(2-\sqrt{3})-4\\\\f(2-\sqrt{3}) = 7-4\sqrt{3}+6-3\sqrt{3}-4\\\\f(2-\sqrt{3}) = 9-7\sqrt{3}$

Therefore, the exact points where the tangent touches f(x) are:

$\Large\boxed{\boxed{\left(2+\sqrt{3},9+7\sqrt{3}\right)\; \textsf{and}\;\left(2-\sqrt{3},9-7\sqrt{3}\right)}}$

$\dotfill$

Additional Notes

To find the equations of the two tangent lines, substitute the x-coordinates of the points of tangency into the slope formula (m = 2x + 3), then substitute the slope and point (2, 3) into the point-slope form of a linear equation.

Tangent line 1

$m=2(2+\sqrt{3})+3=7+2\sqrt{3}$

$y-3=(7+2\sqrt{3})(x-2)\\\\y-3=(7+2\sqrt{3})x-14-4\sqrt{3}\\\\\boxed{y=(7+2\sqrt{3})x-11-4\sqrt{3}}$

Tangent line 2

$m=2(2-\sqrt{3})+3=7-2\sqrt{3}$

$y-3=(7-2\sqrt{3})(x-2)\\\\y-3=(7-2\sqrt{3})x-14+4\sqrt{3}\\\\\boxed{y=(7-2\sqrt{3})x-11+4\sqrt{3}}$