Question

Help thankss
answer is

`( - 14.0)`
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=-(1)/(f'(x_0))(x-x_0)\\\\\\ y-(-4)=-(1)/(4)(x-2)\\\\\\y+4=-(1)/(4)x+(1)/(2)\\\\\\y=-(1)/(4)x+(1)/(2)-4\\\\\\\boxed{y=-(1)/(4)x-(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=-(1)/(4)x-(7)/(2)\\\\\\ 0=-(1)/(4)x-(7)/(2)\\\\\\(1)/(4)x=-(7)/(2)\quad|\cdot4\\\\\\x=-(7\cdot4)/(2)\\\\\\x=-7\cdot2\\\\\\\boxed{x=-14}\\\\\\\boxed{A=(-14,0)}`