Math Revision - Quadratics & Misc. Graphs

How do you find the vertex of a parabola using the axis of symmetry (midpoint)

You substitute the midpoint into the original equation (ax² + bx + c) for x to find the y-coordinate of the vertex

What part of the quadratic equation must be negative to make a parabola concave down

ax²

What do you do when the ax² is actually a number other than 1 or -1 (i.e. 3x² or 7x²)

Divide everything by the coefficient so that x² becomes monic

How do you complete the square?

Halve the x coefficient, then square it, and add it on to the equation (after that, either subtract it from the same line of the equation OR add it to the other side - this makes the equations the same as the original)

What does this type of equation show:

a(x-h)² + k

The vertex/ turning point - where -h is the x-coordinate AND k is the y-coordinate

What is the discriminant?

The b² - 4ac that is inside the square root on the quadratic equation

What happens if the discriminant is 0 ?

The parabola bounces off the x-int rather than going through

Equation of a circle?

x² + y² = r² (radius²)

Equation of a semi circle?

y = square root(a² - x²)

Equation for a hyperbola?

y = 1/x (1 over x) - could be any number not just 1

How do you determine the turning point of a hyperbola

Find the easiest two terms that make the numerator and those are the coordinates (create negative and positive coords) i.e. numerator of 4 —> 2×2 AND -2×-2 (coordinates for turning points would be (2,2) and (-2,-2))

What are the four types of relations and what do they mean?

one-to-one: no x-coords or y-coords match at all - a straight diagonal line

one-to-many: fails the vertical line test

many-to-one: fails the horizontal line test

many-to-many: fails both tests (a cycle)