In general, to plot a parabola, it would help to know the vertex, the zeroes (if any) and the y-intercept.
The y-intercept is simply f(0), i.e. when x=0. For this case f(0)=-5.
The zeroes can be found as
f(x)=x²-5=0
=>
x=±√(5/2)
These are the points where the parabola intersect the x-axis.
The third and most important point is the vertex (h,k) for a parabola in the canonical form:
f(x)=a(x-h)+k
The given parabola is already in the canonical form, from which we determine that h=0, k=-5, or the vertex is at (0,-5).
Finally, the curvature is determined by the leading coefficient a, i.e. the coefficient of x², or 2.
Since a>0, the parabola is concave up. Conversely, if a<0, then the parabola is concave down.
What are three points that you can plot to graph the function f(x)=2x²- 5?
I don't understand this
1 answer
1 answer