find the vertex,focus, direction of the parabola. use a graphing utility to graphthe parabola:
1) x^2-2x+8y+9=0
2) y^2-4y-4y=0
3) x=1/4(y^2+2y+33)
4)(x+1/2)^2=4(y-1)
5) y= 1/4(x^2-2x+5)
please answer my question i need this hw for tomorow
not calculus, algebra
completing the square
I will do one
x^2 -2 x = -2y - 9
note, big x is negative y so opens down, sheds water
-2/2 = -1
(-1)^2 = 1
so add 1 to both sides
x^2 - 2 x + 1 = -2y - 8
(x-1)^2 = -2 (y+4)
vertex at (1,-4)
4a = -2
a = -1/2
so focus at -4.5
this is question no what?
To find the vertex, focus, and direction of a parabola, we need to rewrite each equation in the standard form:
1) x^2 - 2x + 8y + 9 = 0
First, we isolate y by moving all terms to one side:
8y = -x^2 + 2x - 9
Next, divide everything by the coefficient of y to get y by itself:
y = (-1/8)x^2 + (1/4)x - (9/8)
Comparing this equation with the standard form, y = ax^2 + bx + c, we can identify the values of a, b, and c:
a = -1/8
b = 1/4
c = -9/8
The vertex form of the equation is:
y - k = a(x - h)^2, where (h, k) is the vertex.
To find the h-value of the vertex, use the formula:
h = -b/2a
Plugging in the values:
h = -(1/4) / (2 * (-1/8))
h = -1/2
To find the k-value of the vertex, substitute h back into the equation:
k = (-1/8)(-1/2)^2 + (1/4)(-1/2) - (9/8)
k = 5/8
Therefore, the vertex of the parabola is (h, k) = (-1/2, 5/8).
To find the focus, use the formula:
c = 1/4a
Plugging in the value of a:
c = 1 / (4 * (-1/8))
c = -2
Therefore, the focus of the parabola is at a distance of -2 units from the vertex along the y-axis.
The direction of the parabola can be determined by the coefficient of the x^2 term. If a is positive, the parabola opens upwards. If a is negative, the parabola opens downwards.
Since the coefficient of the x^2 term, a, is -1/8 (negative), the parabola opens downwards.
Now, let's use a graphing utility to graph the parabola.