Help with math?

Question

I need help with this problem.
Fine the vertex and intercepts
x = y^2 - 4y + 3

I know how to do it .. but my answer always turns out to be different from the book's answer. Since this equation is a parabola, and given that k in the vertex formula is equal to -b/2a, shouldn't the vertex be at (15,-2) ?

I know what the answer is .. 'cause it's at the back of my book, but how would you go about to find the vertex and intercepts?

Answer 1

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

Vertex is (-1, 2)

when y=0,
x = 3
the parabola intercept x -axis at (3,0)

when x=0,
y^2 - 4y + 3 = 0
(y - 3)(y - 1) = 0
y = 3 or 1
the parabola intercept y -axis at (0,1) and (0,3)

Answer 2

so you have x=y^2-4y+3
you get: x-3=y^2-4y
x-3+4=(y^2-4y+4)
x+1=(y-2)^2
so the vertex is (-1,2)
or if k=-b/2a
b=-4 so k=2

for the intercepts
factor x=(y-3)(y-1)
so y=3,1 when x is 0

Answer 3

15=(-2)^2-4*(-2)+3
15=4+8+3
15=15
the point (15,-2) is on the parabola

intercepts
if y=0
x=3
(3,0)

if x=0
y^2-4y+3=0

y1=1
y2=3

(0,1), (0,3)

Answer 4

Hi,

The vertex is at (-1,2)

The y intercepts are at (0,1) and (0,3)

The x intercept is at (3,0)

Graph y = x^2 - 4x + 3 and reverse your coordinates for this equation.

I hope that helps!! :-)