Find the gradient and equation of the normal to the curve $y=x^3 + x + 5$ at the point where $x=0$ I was wondering if anyone can help with this question.I'm just not sure how to find the gradients and equations of these questions. If you could help it would be greatly appreciated. 
$$y=x^3+x+5$$
 A: by putting $x=0$ to the equation, you get $y=5$ then the point wich you have ask to find the gradient is $(0,5)$
by differentiating the curve equation, you get $\frac{dy}{dx}=3x^2+1$
$(\frac{dy}{dx})_{(x=0,y=5)}=3*0^2+1=1$
curves' gradient at the point $(0,5)$ is $1$
to find the equation of normal to the curve at $x=0$ you have to find the gradient of normal. since the normal and the gradient is perpendicular to each other gradient of normal is $-1$ by (grd(normal)*grd(curve)=-1)
$\frac{y-5}{x-0}=-1$
$y=5-x$ is the equation of normal
A: Hint : First calculate the point $P(0/f(0))$. The derivate of $y=f(x)=x^3+x+5$ at $x=0$ is the slope of the tangent in $P(0)/f(0))$. To get the normal, calculate $m=\frac{-1}{slope\ of\ the\ tangent}$ and use that the point $P(0/f(0))$ is a point of the normal as well.
A: Make your life easier by writing $f(x)=y$ such that 
$$f(x)=x^3+x+5$$
therefore 
$$f(0)=(0)^3+(0)+5=5$$
then we wish to calculate the first derivative with respect to $x$ which is $$f^{\prime}(x)=3x^2+1$$ 
then $$f^{\prime}(0)=3(0)^2+1=1$$
Therefore $\color{blue}{\fbox{$1$}}$ is the gradient of $f(x)$ at $(0,5)$
Now we use the negative reciprocal relation: gradient of normal $= -\cfrac{1}{\text{gradient of tangent}}$ which relates the the gradient of the normal to the gradient of the tangent at $(0,5)$:
Hence, gradient of normal is $-\cfrac{1}{f^{\prime}(0)}=-\cfrac{1}{1}=-1$ 
Now that we have the gradient of the normal we can easily find the equation of the normal by using the general equation for a straight line, namely 
$$\color{#180}{\fbox{$y-y_1=m(x-x_1)$}}$$
where $m$ is the gradient of the straight line and $(x_1,y_1)$ is a point that lies on that line.
So in this specific case $m=-1$ and $(x_1,y_1)=(0,5)$ 
Hence, 
$$y-y_1=\text{gradient of normal}(x-x_1)$$
becomes 
$$y-5=-1(x-0)$$
or 
$$\color{blue}{\fbox{$y=5-x$}}$$
For clarity, the answers to your questions are marked in $\color{blue}{\mathrm{blue}}$.
Here is a graph of your situation showing the curve in $\color{red}{\mathrm{red}}$ and normal to it at the point $(0,5)$:

I tried to answer this question in the most intuitive way possible, and the formula boxed in $\color{#180}{\mathrm{green}}$ is your friend and is easier to find equations of straight lines than using $y=mx+c$ as with that approach you have to substitute a known point to find the intercept $c$ which normally takes longer. 
