SOLUTION: You have a parable with the points -2,0 and 4,0 and a vertex of 0,-3 put into y=a(x-h)^2 +k

SOLUTION: You have a parable with the points -2,0 and 4,0 and a vertex of 0,-3 put into y=a(x-h)^2 +k

Algebra.Com

Question 1133407: You have a parable with the points -2,0 and 4,0 and a vertex of 0,-3 put into y=a(x-h)^2 +k
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
parable?
RELATED QUESTIONS
With a y-intercept 10, x-intercept 2, and equation of axis of symmetry x-3 = 0, find the... (answered by ewatrrr)

Write the equation of the parabola in both vertex and standard form whose vertex is... (answered by ikleyn,josgarithmetic)

Use the quadratic formula to determine the x-intercepts (if any) of the following... (answered by Earlsdon)

concept check for f(x) = a(x-h)^2+k, in what quadrant is the vertex if (a) h>0, k>0; (b)... (answered by josgarithmetic)

find the x intercepts of the parabola with vertex (-1,-16) and y intercept (0,-15). use... (answered by Fombitz,Boreal)

Put the function 2x^2+12x+3 in vertex form and state the values of a,h, and k. vertex... (answered by josgarithmetic)

A parabola of the form y=ax^2+bx+c has a maximum value of y=3. The y-coordinate of the... (answered by josgarithmetic)

Is this correct: What is the equation of a parabola with vertex (0,0) and focus (0,4)? (answered by solver91311,Edwin McCravy)

write the equation y=a(x-h)^2+k with the given. with a y-intercept 10, x-intercept 2,... (answered by stanbon,solver91311)