Question

Help thankss answer is ( - 14.0)" align="absmiddle" class="latex-formula">
the dot is a comma btw

Answer 1

We have point $P=(x_0,y_0)=(2,-4)$, so first calculate $f'(x_0)$. There will be:

$y=f(x)=2x^3-5x^2\\\\\\\\f'(x)=\big(2x^3-5x^2\big)'=\big(2x^3\big)'-\big(5x^2\big)'=2\big(x^3\big)'-5\big(x^2\big)'=\\\\=2\cdot3x^2-5\cdot2x=\boxed{6x^2-10x}\\\\\\\\f'(x_0)=f'(2)=6\cdot2^2-10\cdot2=6\cdot4-20=24-20=\boxed{4}$

Now, we can write the equation of the normal line as:

$y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\ y-(-4)=-\dfrac{1}{4}(x-2)\\\\\\y+4=-\dfrac{1}{4}x+\dfrac{1}{2}\\\\\\y=-\dfrac{1}{4}x+\dfrac{1}{2}-4\\\\\\\boxed{y=-\dfrac{1}{4}x-\dfrac{7}{2}}$

Normal line (and every line) crosses x-axis when y = 0, so coordinates of A:

$\boxed{y=0}\qquad\qquad\text{and}\\\\\\y=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\ 0=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\\dfrac{1}{4}x=-\dfrac{7}{2}\quad|\cdot4\\\\\\x=-\dfrac{7\cdot4}{2}\\\\\\x=-7\cdot2\\\\\\\boxed{x=-14}\\\\\\\boxed{A=(-14,0)}$