whats the vertex of 4y^2+4y-16x+13=0?
rearrange, then complete the square ....
16x = 4y^2 + 4y + 13
4x = y^2 + y + 1/4 - 1/4 + 13/4
4x = (y + 1/2)^2 -1/4 + 13/4
4x = (y+1/2)^2 - 3
x = (1/4)(y + 1/2)^2 - 3/4
can you read the vertex from that?
so is it (-3/4,-1/2)?
yes, good for you to notice that you had to read the x and y in inverse form.
To find the vertex of the quadratic equation 4y^2 + 4y - 16x + 13 = 0, we can rearrange the equation into the standard form of a quadratic equation (y = ax^2 + bx + c). Let's do that:
4y^2 + 4y - 16x + 13 = 0
Rearranging terms:
4y^2 + 4y = 16x - 13
Dividing every term by 4 to make the coefficient of y^2 equal to 1:
y^2 + y = 4x - 13/4
Completing the square by adding (b/2)^2 to both sides, where b is the coefficient of y:
y^2 + y + (1/2)^2 = 4x - 13/4 + (1/2)^2
Simplifying:
y^2 + y + 1/4 = 4x - 13/4 + 1/4
y^2 + y + 1/4 = 4x - 3/2
Now, we can express the equation as a perfect square:
(y + 1/2)^2 = 4x - 3/2
Comparing this with the standard vertex form equation (y - k)^2 = 4a(x - h), we can see that the vertex in this equation is (h, k). Therefore, in our equation:
h = 0 (since there is no x term)
k = -1/2
Hence, the vertex of the equation 4y^2 + 4y - 16x + 13 = 0 is (0, -1/2).