Mishita Dharia - Quadratics
Finding X and Y Intercepts in Vertex Form
Y-intercept: The point at which a parabola crosses the y-axis and when x=0. It would be written as (0, y).
Finding the y-int: On a graph, the y-intercept marks the point where x=0, so in order to find the y-int of a graph, all you have to do is sub 0 in for x in the vertex form of the parabola.
Let’s do an example…
We have the equation y = 2 (x - 3)^2 + 8.
To find the y-int we would have to set x=0…
y = 2(0 - 3)^2 + 8
y = 2(-3)^2 + 8
y = 2(9) + 8
y = 26
y-int: (0, 26)
Now you try...
Find the y-int for -3(x+2)^2+20
Answer...
y-int: (0,8)
X-intercept: A parabola can have zero, one, or two x-ints. These are the points at which the parabola crosses the x-axis and when y=0. It would be written as (x, 0).
Finding the x-ints: On a graph, the x-intercepts mark the points at which y=0, so in order find the x-ints of a graph, you would have to first sub y=0 into the equation and then use the square rot principal to solve for the x-ints.
Let’s do an example…
Find the x-ints of y= 2(x-5)^2-50…
The Square Root Principal: This principal states that when squaring a number, it can have two answers, a positive answer and a negative answer.
For example: 6x6 = 36
Similarly (-6)x(-6) = 36
So…
36 = 6 and (-6)
This is why x = ± a number
There are two possible solutions…
x = + 25 + 5
x = 5 + 5
x = 10
x = 25 + 5
x = (-5) + 5
x = 0
To find the x-int we have to set y=0 and then use the square principal to solve…
0 = 2(x-5)^2-50
50/2=(x-5)^2
50/2 = [2(x-5)^2]/2
25 = (x-5)^2
± 25 = x-5
± 25 + 5 = x
x ints: (10, 0) & (0,0)
Now you try...
Find the x ints for y= -3(x + 2) + 27
Answer...
y-ix-ints: (1, 0) (-5, 0)
Extra Practice...