The picture shows the graph of
$$ y = \frac{1}{10} (x^3 + 3x + 20). $$
I was then told to draw a straight line on the graph, then solve the equation $x^3 + 23x = 30$.
I'm a little confused here about where should I draw the straight line. Can I get help... Thanks!
I suspect the idea is to solve the second equation "graphically" by inserting a line which intersects the first equation at the point $(x, y)$ where $x$ is the solution to the second equation. That is, insert the line $$y = ax + b$$
where $$y=\frac{1}{10}(x^3+3x+20) = ax + b$$
and find the values of $a$ and $b$ such that the above equation becomes the second equation. I.e.$$\frac{1}{10}(x^3+3x+20) = ax + b$$ which means $$x^3+(3-10a)x=10b-20$$
This equation becomes the second equation for the values $$a = -2, b=5$$