We can get points on the parabola, say these 6,
by substituting x=-4, x=-3, x=-2, x=-1, x=0, x=1
(-4,3), (-3,-1), (-2,-3), (-1,-3), (0,-1), (1,3),
We sketch in the parabola:
We want to know where this parabola crosses the x-axis, which I
have marked below with red circles:
From the graph it looks like the x-intercept on the left crosses somewhere
between -3 and -4, and the x-intercept on the right crosses somewhere
between 0 and 1.
To get these accurately we use the quadratic formula to find the values
of x that cause y to be 0:
The solution to
ax² + bx + c = 0
is found by:
Your problem is
x² + 3x - 1 = 0
So a=1, b=3, c=-1
Use a calculator with the +
= 0.3027756377, which is the x-coordinate
of the point circled on the right, and it is between 0 and 1. So
that x-intercept on the right is
(0.3027756377, 0)
Use a calculator with the -
= -3.302775638, which is the x-coordinate
of the point circled on the left, and it is between -3 and -4. So
that x-intercept on the left is
(-3.302775638, 0)
Or you can write them in radical (exact) form:
(
, 0) and (
, 0)
Edwin