- #1
rowdy3
- 33
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Find the equation of the parabola (in standard form) whose vertex is at the origin, has a vertical axis, and passes through the point (-2,5). Thanks
y = ax^2
sub in the point (-2, 5) and solve for a
5 = a(-2)^2
5 = a(4)
a = 5/4
Final equation is y = (5/4)x^2. Is that right?
y = ax^2
sub in the point (-2, 5) and solve for a
5 = a(-2)^2
5 = a(4)
a = 5/4
Final equation is y = (5/4)x^2. Is that right?