Welcome to graphing with calculus. All the other methods you have learned to graph were quite difficult to use when sketching, such as using the multiplicity of the roots to see which intervals increase and decrease.
The method I am going to show is much more accurate than what you learned in the early levels. But first, we need to identify the curve so that we know what kind of function we are graphing. So we want to solve for y.
Subtract 10 and add 5y2 on both sides of the equation. Then subtract 3x2 on both sides of the equation.
5y2 - 10y = -3x2
Divide both sides of the equation by 5.
y2 - 2y = -3x2 / 5
Complete the square. Add 1 on both sides of the equation.
y2 - 2y + 1 = (-3x2 / 5) + 1
(y - 1)(y - 1) = (-3x2 / 5) + 1
(y - 1)2 = (-3x2 / 5) + 1
Square-root both sides of the equation.
y - 1 = ±√((-3x2 / 5) + 1)
Add 1 to both sides of the equation.
y = 1 ± √((-3x2 / 5) + 1)
This equation is a square-root curve.
Now that we have the function of y in terms of x, we can find the x-intercepts. Set y=0 to find them. It will be easier to refer to the original form of the function.
3x2 = 10(0) - 5(0)
3x2 = 0
x = 0
x-intercept is (0, 0).
To find the y-intercept, we set x=0.
0 = 10y - 5y2
0 = 5y(2 - y)
2 - y = 0 and y = 0
y = 2
We have two y-intercepts. (0, 0) and (0, 2).
Next is to find the minimum points and maximum points. We use the first derivative test. We take the first derivative of the function and set is equal to zero. This is because the slope of the line tangent to these points is zero. Since the function can be broken down into two parts, due to the plus and minus sign in the new form of the function, we do this derivative test twice. Once for the positive square-root and again for the negative square-root. I will do the positive part so that you will understand the procedure.
y' = 0
((-3/5)x2 + 1)1/2 = 0
(1/2)((-3/5)x2 + 1)-1/2 * (-6/5)x = 0
Solve for x. When you solve for x, these x values are your critical points. They are the location of the minimum and maximum points.
Then you want to perform test points using these values by evaluating them into the derivative. When the derivative goes from negative to positive, it indicates a minimum point. When the derivative goes from positive to negative, it indicates a maximum point. Once you figure out all of that, evaluate the critical points that have changing derivative signs into the positive y function.
Do the same to the negative square-root part of the function.
Then we find the concavity. This determines the shape of the graph. To do this, we set the second derivative equal to zero.
y'' = 0
When you solve for the x values here, these x values are your points of inflection. Points of inflection indicates the graph changing concavity. You use the test points, just like you did in first derivative test and evaluate the second derivative. When the second derivative is positive, the graph is concave up. When the second derivative is negative, the graph is concave down.
You will need to do this test twice. Once for the positive square-root function, and another for the negative square-root function.
Now that I have given you the info and procedures, can you find all these points?
