Question 46591
f(x) = 2x^2 + x - 1
vertex(-b/2a,f(x))
v(-1/4,-9/8)
x-intercept:
0 = 2x^2 + x - 1
1 = 2x^2 + x
1/2 = x^2 + (1/2)x
8/16 + 1/16 = (x + (1/4))^2
+- sqrt(9/16) = x + 1/4
-1/4 +- 3/4 = x
(-1,0) and (1/2,0)
y-intercept:
f(x) = 2x^2 + x - 1 = 2(0)^2 + 0 - 1 = -1
(0,-1)
Axis of Symmetry:
Since the parabola is vertical, the axis of symmetry is valued as 'x'. {{{x = -1/4}}}
{{{graph(600,600,-10,10,-10,10,2x^2+x-1)}}}