Math with Dhruv


Introduction to Quadratic realaions

Quadratic formulas are used to calculate the height of falling rocks, kicked balls or an arched bridge. The formula is always y=ax^2+bx+c with a = 0. b and c can be 0.

y=3x^2 is an example of b and c as 0. y=2x^2+4x+4 is an example of y=ax^2+bx+c.

Non examples are: y = 5x^3– 5x^2 (2 is not the highest exponent)

y = 3x + 7^2 (can be simplified to 3x + 49, which is linear)

First and Second differences

You will use First and Second difference to determine weather the relation is linear, quadratic or neither. You will take the first number and subtract it by the second number. That will give you one first difference. Then you will take the second number and subtract it by the third number. That will give you another first difference. Keep doing that until all the numbers finish. Then if the first differences are constant the relation is linear. If the first differences are non constant do the same thing with them. And determine the second differences. If the second differences are constant, the relation is quadratic. If both the first and second differences are non constant, then the relation is neither.

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characteristics of Parabola

A Parabola is a graph of a quadratic equation. The line of symmetry for y =ax^2 is the y-axis. The line of symmetry for x=ay^2 is the x-axis. The y-axis is a reflection. suppose there is a point (-5,4), if I wanted to reflect that point on the y-axis the point would be (5,4). (-5,4) would reflect to (-5,-4) on the x-axis

Transformations of Parabola

when you take a Parabola and move it, it is called Transformations of Parabola. You can move the Parabola vertically or horizontally. Suppose your parabola is y=x^2-2 and you want to make it y=x^2+5. This is moving it vertically up 5 units. If you have an equation like y=(x+2)^2 and it is moved to y=(x-1)^2, it is moved 3 units to the left. If you want to move horizontally negative units then the Parabola will shift to the left, and it will shit to the right if you move positive units horizontally.

Writing equations from given transformations

Writing equations from given transformations is just writing the difference in the Parabola. suppose you have a parabola with the equation y=(x+4), and you have to translate it vertically 5 units up and horizontally 3 units to the left. Your Parabola will move 3 units to the left because it is horizontality being moved. And 5 units up. The new Parabola equation will be y=(x+1)^2+5. If you want to reflect Parabola on the x-axis you will just have to add "-" in front of the (x+1).

Determine the equation from given vertex and a point

suppose you have the vertex, (1,3) and the point (4,8). I will show you how to determine the equation from the given information. Parabola of standard form y=a(x-h)^2+k. The vertex is (1,3) which is (h,k) and the giving point (4,8) which is (x,y). Now lets sub in the numbers. 8=a(4-1)^2+3




Find x and y intercepts

If you have to find the x and y intercepts of an equation, you must sub 0 in. If you have to find the x intercept of this equation 3x + 4y = 12 you must sub in 0 for y and solve for x. 3x + 4(0) = 12. 3x = 12, x=12/3, x=4. For y intercept you have to sub 0 for x and solve for y. 3(0) + 4y = 12. 4y=12, y=12/4, y=3. Therefore the x intercept is 4 and the y intercept is 3