How do you find the vertex and intercepts for 4x - y² + 6y - 1 = 0?

1 Answer
May 15, 2016

color(blue)(y_("intercepts")~~5.828" and "0.172" to 3 decimal places")
color(blue)( x_("intercept")=1/4)
color(blue)("Vertex "->(x,y)->(-2,3)

Explanation:

The structure of this is such that it implies f(y) instead of f(x) so that is how I am going to treat it. The consequence is that it rotates an f(x) graph so that the axis of symmetry is parallel to the x-axis instead of the y-axis.

Write as:
" "4x=y^2-6y+1

Divide both sides by 4

" "x=1/4 y^2-6/4y+1/4

color(blue)("Determine the vertex")

Write as:

" "x=1/4(y^2-6y)+1/4

y_("vertex")=(-1/2)xx(-6) = +3

x_("vertex")=1/4(3)^2-3/2(3)+1/4" "=" "-2

color(blue)("Vertex "->(x,y)->(-2,3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine "x" intercept")

Set y=0

color(blue)(=> x_("intercept")=1/4(0)^2-3/2(0)+1/4)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine "y" intercept")

Set x=0

=>1/4(y^2-6y)+1/4=0

color(brown)("Completing the square")

1/4(y-3)^2+1/4 +k=0

Where k is the constant of correction.

k=(-1)xx[1/4(-3)^2] = -9/4 giving:

color(brown)(1/4(y-3)^2+1/4 +k=0)color(blue)(" "->" "1/4(y-3)^2-2=0
'....................................................................

=>(y-3)^2=8

y-3=+-sqrt(8)" "=+-2sqrt(2)

y_("intercept")=3+-2sqrt(2)

color(blue)(y_("intercepts")~~5.828" and "0.172" to 3 decimal places")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B