VCE Maths questions help (1 Viewer)

[boredsatan]

Consider cases of a > 0 and a < 0.

no x intercept means that b^2-4ac < 0
so, 0^2 - 4(p)(q)<0
so, -4pq<0
so, pq <0
is this right?

 

[InteGrand]

no x intercept means that b^2-4ac < 0
so, 0^2 - 4(p)(q)<0
so, -4pq<0
so, pq <0
is this right?

You don't need to do it that way. Note that the condition pq > 0 gives us precisely two possibilities: either p > 0 and q > 0; or p < 0 and q < 0.

If p > 0 and q > 0, we have px^2 + q > 0 for all real x (since x^2 is non-negative for real x), so the quadratic has no real roots.

If p < 0 and q < 0, we have px^2 + q < 0 for all real x (since x^2 is non-negative for real x), so the quadratic has no real roots.

 

[boredsatan]

show that y = 4-6x+2x^2+3x^3 has exactly one x intercept
factorise it
(x+2)(3x^2-4x+2)
b^2 - 4ac = 0
16-4(3)(2) =
16-24 = -8
Not sure how to finish it

anyone?

 

[InteGrand]

show that y = 4-6x+2x^2+3x^3 has exactly one x intercept
factorise it
(x+2)(3x^2-4x+2)
b^2 - 4ac = 0
16-4(3)(2) =
16-24 = -8
Not sure how to finish it

anyone?

What would be the x-intercepts, given your factorisation?

 

[boredsatan]

use algebra to simplify
(x^4 - 16/x+2)/(x^2+4/x-2)
so, ((x^2-4)(x^2+4))/(x+2) * (x-2)/(x^2+4)
not sure how to finish it

 

[boredsatan]

What would be the x-intercepts, given your factorisation?

x = -2, and the other factor doesn't doesn't have an x intercept

 

[boredofstudiesuser1]

show that y = 4-6x+2x^2+3x^3 has exactly one x intercept
factorise it
(x+2)(3x^2-4x+2)
b^2 - 4ac = 0
16-4(3)(2) =
16-24 = -8
Not sure how to finish it

Δ < 0 for 3x^2-4x+2
Therefore no real roots

When x+2 = 0
x = -2

Only x-intercept is x=-2

Therefore y only has exactly one x-intercept

 

[boredsatan]

A model for the density of a koala population is given by
D = (9t+20)/(t+1), t greater than or equal to 0
show that D = 9+ 11/t+1
not sure what the question is asking

 

[boredsatan]

use algebra to simplify
(x^4 - 16/x+2)/(x^2+4/x-2)
so, ((x^2-4)(x^2+4))/(x+2) * (x-2)/(x^2+4)
not sure how to finish it

anyone?

 

[boredsatan]

y = (x-2)^2 + 1
inverse function
x = (y-2)^2 + 1
(y-2)^2 + 1 = x
(y-2)^2 = x - 1
y - 2 = √(x-1)
y = √(x-1) + 2
is this correct?

 

[boredsatan]

f(x) = (x-3)^2 (x+2)
find x: f(x) >6
what does the question mean?

 

[InteGrand]

f(x) = (x-3)^2 (x+2)
find x: f(x) >6
what does the question mean?

Find all values of x such that f(x) > 6.

 

[boredsatan]

(x-3)^2(x+2) >6
so, (x-3)^2(x+2) =6
so, (x-3)^2(x+2) - 6 = 0
is this correct so far?

 

[boredsatan]

(x-3)^2(x+2) >6
so, (x-3)^2(x+2) =6
so, (x-3)^2(x+2) - 6 = 0
is this correct so far?

anyone?

 

[Squar3root]

(x-3)^2(x+2) >6
so, (x-3)^2(x+2) >6
so, (x-3)^2(x+2) - 6 > 0
is this correct so far?

yeah it's correct but keep the inequality signs, i have fixed it above

once you solve that inequality you will see that the roots are +sqrt(3), -sqrt(3) and 4

and if u draw the graph the inequality satisfies when -sqrt(3) < x < +sqrt(3) and x > 4

 

[boredsatan]

Consider the curve with equation y = 2x^2 - x
a. express the gradient of the secant through the points on the curve where x = -1 and x = -1+ h in terms of h
b. Use h = 0.01 to obtain an estimate of the gradient of the tangent to the curve at x = -1
c. deduce the gradient of the tangent to the curve at the point where x = -1
no idea how to do part b and c

 

[InteGrand]

Consider the curve with equation y = 2x^2 - x
a. express the gradient of the secant through the points on the curve where x = -1 and x = -1+ h in terms of h
b. Use h = 0.01 to obtain an estimate of the gradient of the tangent to the curve at x = -1
c. deduce the gradient of the tangent to the curve at the point where x = -1
no idea how to do part b and c

Once you've done a), b) is easy (just put h = 0.01 in your formula from a)). To get the answer to c), take the limit as h -> 0 in your answer for part a).

 

[boredsatan]

(1√6 - 2√9+3√4)/(9√4-4√9)
How would this be simplified?

 

[boredsatan]

(1√6 - 2√9+3√4)/(9√4-4√9)
How would this be simplified?

Anyone?

 

[kawaiipotato]

(1√6 - 2√9+3√4)/(9√4-4√9)
How would this be simplified?

sqrt(9) = 3,
sqrt(4) = 2.
The fraction will simplify to (sqrt(6))/(6).