# How can I find the zeros , x value and y value of these 2 equations: y= -1/7 (x^2-49) and y = -x^2+4x-3

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y=-1/7(x^2-49

To find the function's zero's means that you need to find x values in which the function (y) equals 0.

Let us set the function to 0

-1/7 (x^2-49) =0

Multiply by -7 both sides:

x^2-49=0

Factorize:

(x-7)(x+7)=0

Then, in order for the function to be zero, either (x-7) or (x+7) must equal zero:

x-7= 0 ==> x= 7

x+7=0 ==> x=-7

Then the function has two solution x={-7,7}

y=-x^2+4x-3

Set the function to zero:

-x^2+4x-3=0

or:

x^2-4x+3=0

(x-3)(x-1)=0

Then x=3

and x= 1

To find the zeros of y(x) = -1/7((x^2-49) and y(x) = -x^2+4x-3.

Solution:

Set y = 0 and solve for x in both equation, to get the zeros of the function y(x)

y = -(1/7)(x^2-49).Put y =0.

0 = (-1/7)(x^2-49). Mutliplying by -7,

0= x^2-49.Swapping,

x^2-49 = 0.

(x+7)(x-7) = 0.

x=-7 and x =7 are the zeros of y(x)

2)

y = -x^2+4x-3. Put y = 0.

0=-x^2+4x+3 .Or

x^2-4x+3 = 0 . Or

x^2-3x -x+3 = 0.

x(x-3)-1(x-3) = 0. Pul out x-2

(x-3)(x-1) = 0.

Equating the facrors to zero, and so;ving for x,

x = 3 Or x = 1