Question

How do I find the x-intercepts of the equation y=-0.25(x+4)^2+1 without graphing

Answer 1

what do you think will be the value of y when the curve intersects the x-axes?

Answer 2

0

Answer 3

But im getting the wrong answer

Answer 4

correct, so, to find the x-intercepts, just set y=0 and solve for x

Answer 5

please show your steps so I can spot where you may have made a mistake

Answer 6

y=-0.25(x+4)^2+1
0=-0.25(x+4)^2+1
-1 = -0.25(x+4)^2
-1/-0.25 = x^2+16

Answer 7

(x+4)^2 is not equal to x^2+16

Answer 8

also, you do not need to expand this

Answer 9

So how can I solve from here?

Answer 10

x + 4 = +-2

Answer 11

you got to:\[\frac{-1}{-0.25}=(x+4)^2\]

Answer 12

now look at the left-hand-side

Answer 13

4 = (x+4)^2

Answer 14

you have a minus at the top and bottom of the fraction

Answer 15

\[(x + 4)^2 = 4 \implies x+4 = \pm 2\]

Answer 16

correct

Answer 17

then just take square roots of both sides

Answer 18

Oh, I have to take the square root away first, got you.
Thanks

Answer 19

that is right - that is why you didn't need to expand (x+4)^2

Answer 20

can you explain the + or - thing waterineyes did?

Answer 21

the square root of 4 is -2 or +2

Answer 22

because (-2)^2 = 4
and (2)^2 = 4

Answer 23

When you take the square root of a number then you put + and - sign in front of it:
Because:
\[(-x)^2 = x^2\]
\[(x)^2= x^2\]

From both -x and x you are getting \(x^2\)..

Answer 24

Thanks alot I never thought I would get the answer

Answer 25

yw :)

Answer 26

So, one value of x you will get by taking +2 and other value you will get by taking -2..

Go ahead,,

Answer 27

-2-4 = -6
2-4 = -2

This marks the two x-intercepts :)

Answer 28

sometimes the parabola only has one x-intercept or none how do I know that

Answer 29

My concepts are not deeper, @asnaseer will help you here..

Answer 30

if the equation has two solutions (as you have here), then it has 2 x-intercepts
if it has one solution, then there is only one solution
if it has no solution then it does not cross the x-axes

Answer 31

how much have you studied quadratic equations?
have you heard about the determinant?

Answer 32

no, not yet

Answer 33

I guess I will find out later on then

Answer 34

I think you will. The general quadratic has the form:\[ax^2+bx+c=0\]it's determinant is given by:\[b^2-4ac\]where a, b and c are constants

Answer 35

if the determinant is positive, then the quadratic has 2 solutions
if the determinant is zero, then the quadratic has 1 solution
and if the determinant is negative, then the quadratic has no solutions

Answer 36

you can visualise this as follows: