How do you find the vertex and intercepts for x – 4y^2 + 16y – 19 = 0?

2 Answers
Jul 26, 2018

Vertex (3,2)
X-intercept (19,0)

Explanation:

x-4y^2+16y-19=0

x-19=4y^2-16y

x-19+16=4(y^2-4y+4)

x-3=4(y-2)^2

(y-2)^2=1/4(x-3) which is in the form (y-k)^2=4a(x-h) where (h,k) is the vertex

Hence (3,2) is the vertex.

For the intercepts,
When y=0
(0-2)^2=1/4(x-3)
4=1/4(x-3)
16=x-3
x=19

When x=0,
(y-2)^2=1/4(0-3)
(y-2)^2=1/4(-3)
(y-2)^2=-3/4
Since you cannot squareroot a negative number, there are no y-intercepts

Below is what the graph looks like

graph{x-4y^2+16y-19=0 [-10, 10, -5, 5]}

Jul 26, 2018

Please see the explanation below.

Explanation:

The equation is

x-4y^2+16y-19=0

4y^2-16y=x-19

y^2-4y=1/4(x-19)

Completing the square

y^2-4y+4=1/4(x-19)+4

Factorising

(y-2)^2=1/4x-19/4+4=1/4x-3/4

The equation of the parabola is

(y-2)^2=1/4(x-3)

Comparing this to the equation of a parabola

(y-b)^2=2p(x-a)

The vertex is V=(a,b)=(3,2)

The intercepts are when y=0

=>, 4=1/4(x-3)

=>, 16=x-3

=>, x=16+3=19

The point is =(19,0)

graph{x-4y^2+16y-19=0 [0.74, 20.74, -3.32, 6.68]}