I am meanwhile doing Paraballas in math, and I need help understanding.
I know how to find the y-intercept, the vertex, and the axis of symetry using a quadratic equation. The only thing I think I am fuzzy on is how to find the TWO x-intercepts. If anyone could help me, it would be much appreciated. Thank you.
2006-10-23 11:42:44 · 2 answers · asked by Jazz 2 in Science & Mathematics ➔ Mathematics
This is an applet to explore the equation of a parabola and its properties. The equation used is the standard equation that has the form
(y - k)2 = 4a(x - h) or y2 - 4y - 4x = 0
where h and k are the x- and y-coordinates of the vertex of the parabola and a is a non zero real number (in this investigation we consider only cases with positive a). For the definition and construction of a parabola.
2006-10-23 11:52:57 · answer #1 · answered by Jamil Ahmad G 3 · 0⤊ 0⤋
If you know how to find the vertex, put the equation into vertex form: y=k(x-a)^2+b and set this equal to zero and solve for the two zeros of x. For example, if you know that y=2(x-3)^2-4
since y=0 at the x-intercepts, 0=2(x-3)^2-4.
This means that 4=2(x-3)^2...2=(x-3)^2....+-sqrt(2)=x-3
finally, +-sqrt(2)+3=x.
2006-10-23 11:47:03 · answer #2 · answered by bruinfan 7 · 0⤊ 0⤋