# Math Help - Graphs, parabolas, hyperbolas...

1. ## Graphs, parabolas, hyperbolas...

Hi.

I'm really stuck on a few questions I have to do. :/
Help would be greatly appreciated.

1. Sketch the graph of: y = (x-2)2 (2x+5) showing all points of interest. The secondary stationary point occurs when x=-1

2. The Parabola y = 2x2 + 5x + 8 and the line y = ax - 10 intersect exactly at one point. Determine value(s) of a and the point of intersection.

3. A truncas has intercepts of (2,0), (6,0), and (0,6). Determine its equation, domain, and range.

4. The remainder when f(x) = 4x4 - 4x3 + jx2 + x + 6 is divided by x-2 is -60. Demonstrate that j must be equal to -25.
Use long division to then factorise f(x)

2. ## Re: Graphs, parabolas, hyperbolas...

Originally Posted by QueenLuv
1. Sketch the graph of: y = (x-2)2 (2x+5) showing all points of interest. The secondary stationary point occurs when x=-1
To find the zeros make y=0 and solve for x. To find the stationary pts make y'=0 and solve for x

Originally Posted by QueenLuv

2. The Parabola y = 2x2 + 5x + 8 and the line y = ax - 10 intersect exactly at one point. Determine value(s) of a and the point of intersection.
Make the equations equal, then solve the discriminant equalling zero

Originally Posted by QueenLuv

3. A truncas has intercepts of (2,0), (6,0), and (0,6). Determine its equation, domain, and range.
You have $y = \frac{a}{(x-b)^2}+c$

Subsitute your values one at a time and solve the system

Originally Posted by QueenLuv

4. The remainder when f(x) = 4x4 - 4x3 + jx2 + x + 6 is divided by x-2 is -60. Demonstrate that j must be equal to -25.
Use long division to then factorise f(x)
Note that f(2) = -60

3. ## Re: Graphs, parabolas, hyperbolas...

Thanks. But I'm still stuck on 2 and 4. The wording of them really confuse me. :/

So sorry, but could you (or anyone else) in layman's terms, please tell me the steps on how to get the answers?

4. ## Re: Graphs, parabolas, hyperbolas...

Hello, QueenLuv!

1. Sketch the graph of: $y \:=\: (x-2)^2(2x+5)$ showing all points of interest.
The secondary stationary point occurs when $x=-1.$

The graph is a cubic with a positive leading coefficient.
. . Its shape is: . $\begin{array}{c}\cap \\ [-3mm]\;\;\;\cup \end{array}$

The graph has x-intercepts $2$ and $-\tfrac{5}{2}$
The graph is tangent to the x-axis at $x = 2.$ (minimum)
The y-intercept is: . $y \,=\,(-2)^2(5) \,=\,20$

When $x = -1,\:y \:=\:(-3)^2(3) \:=\:27$ (maximum)

Can you sketch the graph now?

5. ## Re: Graphs, parabolas, hyperbolas...

Hello again, QueenLuv!

4. The remainder when $f(x) \:=\: 4x^4 - 4x^3+ jx^2 + x + 6$ is divided by $x-2$ is $-60.$
Demonstrate that $j$ must be equal to $-25.$

Use long division to then factorise $f(x)$.

We are told that: . $f(2) \,=\,-60$

. . $4(2^4) - 4(2^3) + j(2^2) + 2 + 6 \:=\:-60 \quad\Rightarrow\quad 4j + 40 \:=\:-60$

. . $4j \:=\:-100 \quad\Rightarrow\quad \boxed{j \:=\:-25}$

$\begin{array}{cccccccccccc} &&&& 4x^3 & - & 12x^2 & - & x & + & 3 \\ && -- & -- & -- & -- & -- & -- & -- & -- & -- \\ x+2 & | & 4x^4 &-& 4x^3 &-& 25x^2 &+& x &+& 6 \\ && 4x^4 &+& 8x^3 \\ && -- & -- & -- \\ &&& - & 12x^3 &-& 25x^2 \\ &&& - & 12x^3 &-& 24x^2 \\ &&& -- & -- & -- & -- \\ &&&&& - & x^2 &+& x \\ &&&&& - & x^2 &-& 2x \\ &&&&& -- & -- & -- & -- \\ &&&&&&&& 3x &+& 6 \\ &&&&&&&& 3x &+& 6 \\ &&&&&&&& -- & -- & -- \end{array}$

$\begin{array}{ccccccccccc} &&&& 4x^2 &&& - & 1 \\ && -- & -- & -- & -- & -- & -- & -- \\ x-3 & | & 4x^3 &-& 12x^2 &-& x &+& 3 \\ && 4x^3 &-& 12x^2 \\ && -- & -- & -- \\ &&&&& - & x &+& 3 \\ &&&&& - & x &+& 3 \\ &&&&& -- & -- & -- & -- \end{array}$

Therefore: . $f(x) \;=\;(x+2)(x-3)(4x^2-1) \;=\;(x+2)(x-3)(2x-1)(2x+1)$

6. ## Re: Graphs, parabolas, hyperbolas...

Oh, now I get it. Thank you so much!!!