# Thread: VCE Maths questions help

1. ## Re: VCE Maths questions help

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

2. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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.

3. ## Re: VCE Maths questions help

(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?

4. ## Re: VCE Maths questions help

Originally Posted by 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?

5. ## Re: VCE Maths questions help

Originally Posted by 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?
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

6. ## Re: VCE Maths questions help

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

7. ## Re: VCE Maths questions help

Originally Posted by 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
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).

8. ## Re: VCE Maths questions help

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

9. ## Re: VCE Maths questions help

Originally Posted by boredsatan
(1√6 - 2√9+3√4)/(9√4-4√9)
How would this be simplified?
Anyone?

10. ## Re: VCE Maths questions help

Originally Posted by boredsatan
(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).

11. ## Re: VCE Maths questions help

is 3/(x-1)+2 a polynomial ? If no, why isn't it a polynomial?

12. ## Re: VCE Maths questions help

Originally Posted by boredsatan
is 3/(x-1)+2 a polynomial ? If no, why isn't it a polynomial?
no

polynomials have integer powers

13. ## Re: VCE Maths questions help

Originally Posted by Squar3root
no

polynomials have integer powers
what are integer powers?

thanks
for the help

15. ## Re: VCE Maths questions help

Originally Posted by boredsatan
what are integer powers?
Powers that are integers (whole numbers)

Also polynomials must have POSITIVE integer powers not just integers, so y = x^-3 isn't a polynomial

16. ## Re: VCE Maths questions help

Originally Posted by boredsatan
what are integer powers?
$\noindent Integer powers are exponents that belong to the set \mathbb{Z} = \{\ldots -3, -2 , -1, 0, 1, 2, 3 \ldots \}. But polynomials only have \textbf{non-negative} integer exponents, \{0, 1, 2, 3, \ldots \}$

17. ## Re: VCE Maths questions help

Originally Posted by boredsatan
is 3/(x-1)+2 a polynomial ? If no, why isn't it a polynomial?
Originally Posted by Squar3root
no

polynomials have integer powers
Originally Posted by boredsatan
what are integer powers?
Originally Posted by pikachu975
Powers that are integers (whole numbers)

Also polynomials must have POSITIVE integer powers not just integers, so y = x^-3 isn't a polynomial
Originally Posted by 1729
$\noindent Integer powers are exponents that belong to the set \mathbb{Z} = \{\ldots -3, -2 , -1, 0, 1, 2, 3 \ldots \}. But polynomials only have \textbf{non-negative} integer exponents, \{0, 1, 2, 3, \ldots \}$

yes this

polynominals have integer powers > 0

1729 mentioned toe subset above

18. ## Re: VCE Maths questions help

How would you show that the points (5,2). (-2,1) and (1,5) form the vertices of an isosceles triangle?

19. ## Re: VCE Maths questions help

all of the points (-1,-6), (0,-6), (1,-2) and (2,2) lie on the cubic with equation ax^3 + bx^2 + cx + d.
Find the value of d
determine 3 simultaneous equations that will enable solutions to a, b, and c to be found
find the values of a, b, and c, and determine the equation of the cubic

20. ## Re: VCE Maths questions help

a graph has the rule
y = a/(x-b)^2 + c
the y intercept is located ta (0,-1) and an asymptote exists at y = 2, determine the values of a, b, and c

21. ## Re: VCE Maths questions help

Would appreciate any help. Have a maths exam in 4 days

22. ## Re: VCE Maths questions help

If f(x) = 3x and g(x) = x+2, then what is the value of g(f(-1))

23. ## Re: VCE Maths questions help

Originally Posted by boredsatan
all of the points (-1,-6), (0,-6), (1,-2) and (2,2) lie on the cubic with equation ax^3 + bx^2 + cx + d.
Find the value of d
determine 3 simultaneous equations that will enable solutions to a, b, and c to be found
find the values of a, b, and c, and determine the equation of the cubic
\noindent The coordinates of the given points must satisfy y = ax^3 + bx^2 + cx + d. \begin{align*}-a + b - c + d &= -6 \quad (1)\\ d &= -6 \quad (2) \\ a + b + c +d &= -2 \quad (3) \\ 8a + 4b + 2c + d &=2 \hspace{3mm} \quad (4) \end{align*} \\ Use (2) to reduce (1), (3) and (4) to a set of three simultaneous equations and solve for a, b and c.

24. ## Re: VCE Maths questions help

Originally Posted by boredsatan
If f(x) = 3x and g(x) = x+2, then what is the value of g(f(-1))
First find f(-1).

Then plug this into the equation for g(x) to get the value of g(f(-1)).

25. ## Re: VCE Maths questions help

Originally Posted by boredsatan
If f(x) = 3x and g(x) = x+2, then what is the value of g(f(-1))
$\noindent f(-1) = 3(-1) = -3 and so g(f(-1)) = g(-3) = (-3) + 2 = -1.$

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