Question

Use the method of completing the square to write the equation of the given parabola in this form:

(y-k)=a(x-h)^2

where a is not equal to 0, (h,k) is the vertex, and x=his the axis of symmetry

Find the x-intercept of the parabola with vertex (1,1) and y-intercept (0,-3). Write your answer in this form: (x1,y1),(x2,y2)

Answer 1

Formula is y = a(x-h)^2 + k

Where h is 1 and k is 1

f (x) = a(x-1)^2 + 1

-3 = a(0-1)^2 + 1

-3 = a(-1)^2 + 1

-3 = a(1) + 1

-3 - 1 = a

-4 = a

a = -4

A must be equal to -4

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

0 = -4(x-1)^2 + 1

4(x^2 - 2x + 1) - 1 = 0

4x^2 - 8x + 4 - 1 = 0

4x^2 - 8x + 3 = 0

4x^2 - 8x = -3

Divide fpr 4 each term of the equation....x^2 - 2x = -3/4

We must factor the perfect square ax^2 + bx + c which we don't have. We must follow the rule (b/2)^2 where b is -2....(-2/2)^2 = (-1)^2 = 1 and we add up that to both sides

x^2 - 2x + 1 = -3/4 + 1

x^2 - 2x + 1 = 1/4

(x-1)^2 = 1/4

square root both sides x-1 = (+/-) 1/2

x1 = +1/2 + 1 = 3/2

x2 = -1/2 + 1 = 1/2

x-intercepts are 1/2 and 3/2, in form (3/2,0); (1/2,0)

Answer 2

So in this question we are trying to find the two x - intercepts.

$y = a(x-h)^2 + k$  
is the formula


$f (x) = a(x-1)^2 + 1$

$-3 = a(-1)^2 + 1$

$-3 = a(1) + 1$

$-3 - 1 = a$

$a = -4$
So a is -4.

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

$4(x^2 - 2x + 1) - 1 = 0$

$4x^2 - 8x + 4 - 1 = 0$

$4x^2 - 8x + 3 = 0$

$4x^2 - 8x = -3$

So our standard form is:
$4x^2 - 8x = -3$

$( \frac{3}{2} ,0)$

and

$(\frac{1}{2} ,0)$

are you x - intercepts.

Also, in Plato put a comma in-between the intercepts

$(x_{1} y_{1}),( x_{2} , y_{2} )$

Hope I've helped!