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functions and relations HELP!!! (2 Viewers)
- Thread starter Mdzabakly
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Carrotsticks
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a) i) -a^2 is always negative, so it will fall into the |2x| category.
ii) Just sketch each component for each given domain.
b) i) Sub in x=0
ii) Taking the limit as x -> infinity, the bottom gets very large so the whole term gets very small. The limit is 0 from the negative side (because of the negative sign there)
iii) f(-x) = f(x) quite clearly since we only have x^2 there, so even function.
iv) Domain is all real x since nothing funky can happen like vertical asymptotes. Range can be found by re-arranging the function to make it a quadratic in terms of x (so y is constant) then letting the discrminant be >= 0.
v) Just put pieces together to sketch it. Looks like an upside-down bell curve.
c) i) Complete the square to do this.
ii) You can obtain the radius directly from the equation once you've completed the square.
iii) Find perpendicular distance between the line and point. If it is larger than the radius, then clearly it doesn't intersect it. If it is equal to the radius, then it is tangential. If it is less than the radius, then it intersects the circle twice.
Alternatively, sub the line into the equation and find the discriminant, and judge how many times it intersects based on that.
ii) Just sketch each component for each given domain.
b) i) Sub in x=0
ii) Taking the limit as x -> infinity, the bottom gets very large so the whole term gets very small. The limit is 0 from the negative side (because of the negative sign there)
iii) f(-x) = f(x) quite clearly since we only have x^2 there, so even function.
iv) Domain is all real x since nothing funky can happen like vertical asymptotes. Range can be found by re-arranging the function to make it a quadratic in terms of x (so y is constant) then letting the discrminant be >= 0.
v) Just put pieces together to sketch it. Looks like an upside-down bell curve.
c) i) Complete the square to do this.
ii) You can obtain the radius directly from the equation once you've completed the square.
iii) Find perpendicular distance between the line and point. If it is larger than the radius, then clearly it doesn't intersect it. If it is equal to the radius, then it is tangential. If it is less than the radius, then it intersects the circle twice.
Alternatively, sub the line into the equation and find the discriminant, and judge how many times it intersects based on that.
Still not understanding how to do C... can i get a more detailed explination
( also thanks already carrot)
(c) The equation of a circle is given by the equation x^2+y^2-2x+4y-11=0
.
(i) Show that the centre is (1,-2).
(ii) Find the length of its radius.
(iii) Does the line 2x - y + 8 = 0 intersect the circle? Explain.
( also thanks already carrot)
(c) The equation of a circle is given by the equation x^2+y^2-2x+4y-11=0
.
(i) Show that the centre is (1,-2).
(ii) Find the length of its radius.
(iii) Does the line 2x - y + 8 = 0 intersect the circle? Explain.
Carrotsticks
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(i) That equation of a circle as it is right now isn't really helpful. We need to put it in the form:
^2 + (y-y_1)^2 = r^2 $ where $ r $ is the radius and $ (x_1,y_1) $ is the centre of the circle.)
Once you've completed the square and done all that, you get part (i) and (ii) immediately.
(iii) Think about it this way... suppose there is a circle of radius 20m. If I am 25m away from the centre, am I inside it? Obviously no. If I am EXACTLY 20m away from the centre, then I am standing on the circumference aren't I? If I am say 15m away from the centre, then of course I am INSIDE the circle.
Same idea but with a line instead of me. But the only way to compute the distance from a line to a point is the use the perpendicular distance formula, so we do exactly that.
Once you've completed the square and done all that, you get part (i) and (ii) immediately.
(iii) Think about it this way... suppose there is a circle of radius 20m. If I am 25m away from the centre, am I inside it? Obviously no. If I am EXACTLY 20m away from the centre, then I am standing on the circumference aren't I? If I am say 15m away from the centre, then of course I am INSIDE the circle.
Same idea but with a line instead of me. But the only way to compute the distance from a line to a point is the use the perpendicular distance formula, so we do exactly that.
umm.. so how does So how does x^2+y^2-2x+4y-11=0 become (x-1)^2 + (y +2)^2 =16
( sorry again im just really sheit at maths =-=)
Also..
Also;
The curve y=ax^3+bx passes through the point (1,7). The tangent at this point is parallel to the line y=2x-6
Find the values of a and b
((Also thank u so much carrot u saved meee))
( sorry again im just really sheit at maths =-=)
Also..
Also;
The curve y=ax^3+bx passes through the point (1,7). The tangent at this point is parallel to the line y=2x-6
Find the values of a and b
((Also thank u so much carrot u saved meee))
Last edited:
Carrotsticks
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No worries, the forum IS for asking questions!
^2 + (y+2)^2 = 16 \\\\ $Therefore the centre is $ (1,-2) $ and the radius is 4.$ )
f(-6) falls under the top category, so f(-6)=-2
f(1) falls under the second category, so f(1)=0
f(6) falls under the last category, so f(6)=6 since f(x)=x.
So therefore the final answer is -2 + 0 + 6 = 4.
(ii) Note that a^2+2 is always greater than or equal to 2 (no matter what value of 'a' I sub in), so it falls under the bottom category. Therefore f(a^2+2) = a^2+2 because f(x)=x.
f(-6) falls under the top category, so f(-6)=-2
f(1) falls under the second category, so f(1)=0
f(6) falls under the last category, so f(6)=6 since f(x)=x.
So therefore the final answer is -2 + 0 + 6 = 4.
(ii) Note that a^2+2 is always greater than or equal to 2 (no matter what value of 'a' I sub in), so it falls under the bottom category. Therefore f(a^2+2) = a^2+2 because f(x)=x.
4025808
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Equate the two together, that means:
2x^2 = 1 - x
Solve for x, you'll get two values of x. Then you substitute those values into y so you can get the values of both.
Sy123
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