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

Question

2102 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-26T08:48:27

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

#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]}

Answer 2 · 2018-07-26T09:11:42

Please see the explanation below.

The equation is

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

#4y^2-16y=x-19#

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

#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]}