To find the points of intersection between the graph of the quadratic equation y = x² and the circle centered at (0, 1) with a radius of 3, we need to set up the equations.
The equation of the circle can be expressed as:
(x – 0)² + (y – 1)² = 3²
Which simplifies to:
x² + (y – 1)² = 9
Now, we substitute the expression for y from the first equation into the circle’s equation:
x² + (x² – 1)² = 9
This expands to:
x² + (x^4 – 2x² + 1) = 9
Combining like terms, we have:
x^4 – x² – 8 = 0
This is a quartic (fourth-degree) polynomial equation. Let’s define a new variable:
u = x²
The equation then becomes:
u² – u – 8 = 0
Now we can solve for u using the quadratic formula:
u = (1 ± √(1 + 32)) / 2
u = (1 ± √33) / 2
Calculating this gives us two values for u. Since u = x², we can see that both solutions will yield positive x values (if interchangeable), but we need to check if they result in real x points:
Taking the square root of each u value gives two possible x values per u:
For each positive value of u, we can extract two values for x (positive and negative). In total, we have four points of intersection when considering both positive and negative roots.
Thus, the original question is answered: the graph of y = x² and the circle centered at (0, 1) with a radius of 3 intersect at four points.