Suppose a parabola has an axis of symmetry at x = -8 , a maximum height of 2, and passes through the point (-7, -1). write the equation of the parabola in vertex form.
Suppose a parabola has an axis of symmetry at x = -8 , a maximum height of 2, and passes through the point (-7, -1). write the equation of the parabola in vertex form.
As it has a maximum value the coefficient of x^2 will be negative
The vertex will be at (-8,2) so in vertex form it is
y = a(x + 8)^2 + 2
and as it passes through (-7,-1) we have:
-1 = a(-7+8)^2 + 2
-1 = a + 2 so a = -3
answer is y = -3(x + 8)^2 + 2
in standard form this is y = -3x^2 -48x - 190
Its standard form of equation: (x-h)^2=-4p (y-k), (h,k)=(x,y) coordinates of the vertex
For given parabola: Vertex = (-8, 2)
(x-(-8))² = -4p(y - 2)
solve for 4p using coordinates of given point ( -7, -1)
-7+8 = 4p (-1 - 2)
1 = 4p(-3)
4p = -1/3
So, Equation of given parabola:
(x+8)² = -(y-2) / 3
Hope this helps!
Y=-3(x+8)^2+2 Fill in the vertex (-8,2), and the ordered pair(-7.1) then solve for a. Y = a(x-h)^2 + k
What if it doesnt have an axis of symmetry?
we know that
If an axis of symmetry is [tex]x=-8[/tex] and has a maximum height of [tex]2[/tex]
then
Is a vertical parabola open down with vertex at [tex](-8,2)[/tex]
the equation in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
substitute
[tex]y=a(x+8)^{2}+2[/tex]
Find the value of a
the parabola passes through the point [tex](-7,-1)[/tex]
substitute in the formula
[tex]-1=a(-7+8)^{2}+2[/tex]
[tex]-1=a(1)^{2}+2[/tex]
[tex]-1=a+2[/tex]
[tex]a=-3[/tex]
therefore
the answer is
the equation in vertex form is [tex]y=-3(x+8)^{2}+2[/tex]