Question

can someone please help me figure out the x and y intercepts of this quadratic equation?
y=2(x+4)^2-3

Answer 1

in the file)

Answer 2

If i get it right, you want y=0 => 2(x+4)^2-3=0 => x=sqrt(3/2)-4 AND x=-sqrt(3/2)-4

and x=0 , so y=2*4^2-3 => y= 29

SO there are three points of intersect, A(sqrt(3/2)-4,0), B(-sqrt(3/2)-4,0) and C(0,29)

Note sqrt is square root :-)

Answer 3

@Vladislav I can't view that file because it says that i need to download something....and I can't :/
can you print screen the what was in the document so that I can see?

Answer 4

ok)

Answer 5

@Theo177 I understood how to find the y-axis but I'm still quite confused for the x-axis
@Vladislav So that's the graph of the equation right?

Answer 6

Yes. x-horizontal, у-vertical

Answer 7

X axis has 2 intercect points.
\[\sqrt{3/2} -4\] is just a number

Answer 8

2(x+4)^2-3 = 2(x^2 +8x +16)-3 = 2x^2 + 16x +29
set = 0
2x^2 +16x + 29=0
devide the twos away
x^2 +8x + 29/2 = 0
(x+4)^2 = -29/2 + 16
x+4)^2 = -29/2 + 32/2
(x+4)^2 = 3/2
(x+4) = +-sqrt(3/2)
x = -4 + sqrt(3/2) and -4 - sqrt(3/2)

Answer 9

-0r+ sqrt(3/2) -4 is two numbers

Answer 10

@zzr0ck3r how did you get the number 16 in the 6th line?

Answer 11

\(y\) intercept is what you get when you replace \(x\) by 0
i get
\[y=2(0+4)^2-3=2\times 16-3=32-3=29\]

Answer 12

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

For x=0, replace x with 0

Answer 13

\(x\) intercept you get by setting equal to zero and solve
\[2(x+4)^2-3=0\]
\[2(x+4)^2=3\]
\[(x+4)^2=\frac{3}{2}\]
\[x+4=\pm\sqrt{\frac{3}{2}}=\pm\frac{\sqrt{6}}{2}\]
\[x=-4\pm\frac{\sqrt{6}}{2}\]

Answer 14

Oh thank you @satellite73 for making that clear! but how did you get the square root of 3/2 into the square root of 6/2?