How do you find the vertex and the intercepts for y=x^2-2x-3?

1 Answer
Ori Kiko
Nov 23, 2017

vertex ➝ (1,-4)

x-ax is ➝ (-1,0) ∪ (3,0)
y-ax is ➝ (0,-3)

Explanation:

The vertex formula is:
x_v=(-b)/(2a)

so,
x_v=(2)/(2*1)=1

we substitute in the function equation,
y_v=f(1)=1^2-2*1-3=-4

so we get the point (1,-4).
To know the x-axis interception points we equal the function to 0.
0=x^2-2x-3

x=(-b+-sqrt(b^2-4ac))/(2a)=(2+-sqrt((-2)^2-4*1*(-3)))/(2*1)

x_1=-1
x_2=3

so we have the points (-1,0) and (3,0).
If we substitute xby 0 in the function equation we get the interception point in the y-axis,
f(0)=0^2-2*0-3=-3

we get the point (0,-3).

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