Question
Write the equation of a parabola, in standard form, that goes through these points:
(0, 3) (1, 4) (-1, -6)
ax^2 + bx + c = y
a * 0^2+b*0+c = 3
a*1^2+b+1+c = 4
a*(-1)^2+b(-1)+c = -6
c = 3
a + b + c = 4
a – b + c = -6
a + b + 3 = 4
a – b + 3 = -6
Graph the parabola above. Indicate the vertex and axis of symmetry.
(the graph has a max and min number of -20 to 20, both x and y)
(0, 3) (1, 4) (-1, -6)
ax^2 + bx + c = y
a * 0^2+b*0+c = 3
a*1^2+b+1+c = 4
a*(-1)^2+b(-1)+c = -6
c = 3
a + b + c = 4
a – b + c = -6
a + b + 3 = 4
a – b + 3 = -6
Graph the parabola above. Indicate the vertex and axis of symmetry.
(the graph has a max and min number of -20 to 20, both x and y)
Answers
Steve
why stop there? subtract the equations and you get
2b = 10
b = 5
so, a = -4
so, the equations is
-4x^2+5x+3 = 0
as always, the axis of symmetry is at x = -b/2a, so it is
x = 5/8
and the vertex is at (5/8,73/16)
you can see this by writing it as
y = -4x^2+5x+3
= -4(x^2 - 5/4 x) + 3
= -4(x^2 - 5/4 x + 25/64) + 3 + 4*25/64
= -4(x - 5/8)^2 + 73/16
2b = 10
b = 5
so, a = -4
so, the equations is
-4x^2+5x+3 = 0
as always, the axis of symmetry is at x = -b/2a, so it is
x = 5/8
and the vertex is at (5/8,73/16)
you can see this by writing it as
y = -4x^2+5x+3
= -4(x^2 - 5/4 x) + 3
= -4(x^2 - 5/4 x + 25/64) + 3 + 4*25/64
= -4(x - 5/8)^2 + 73/16