Question

Expand and simplify

a) (x+3)(x+6)         b) (x + 6)^2

Given the quadratic relation y = x^2 + 24 - 15

find the maximum or minimum value or the quadratic relation y = *7 + 8x + 15

Find the zeros of the quadratic relation y = x2 + 7x - 18

The curve formed by a rope bridge can be modeled by the relation y = x2 - 11x + 10, where "x" is the horizontal distance in metres and "y" is the height in metres. What is the horizontal distance from one end of the bridge to the other end?

Please show how you got the answer and make it is right

Answer 1

1.

a)

$$(x+3)(x+6)=x^2+6x+3x+18$$

$$=x^2+9x+18$$

b)

$$(x + 6)^2=x^2+2x(6)+(6)^2$$

$$=x^2+12x+36$$

2.

$$y = x^2 + 8x +15$$

$$=x^2+2x(12)+(12)^2-(12)^2-15$$

$$=(x+12)^2-159$$

The minimum value is $-159$ at $x=-12.$

Vertex $(-12, -159).$

The graph opens up.

3.

$$y = x^2 + 24x - 15$$

$$=x^2+2x(4)+(4)^2-(4)^2+15$$

$$=(x+4)^2-1$$

The minimum value is $-1$ at $x=-4.$

Vertex $(-4, -1).$

The graph opens up.

4.

$$y = x^2 + 7x -18$$

$$x^2 + 7x -18=0$$

$$D=(7)^2-4(1)(-18)=121>0$$

$$x=\dfrac{-7\pm\sqrt{121}}{2(1)}=\dfrac{-7\pm11}{2}$$

$$x_1=\dfrac{-7-11}{2}=-9, x_2==\dfrac{-7+11}{2}=2$$

$\{-9,2\}$

5.

$$y=x^2-11x+10$$

$$x^2 -11x+10=0$$

$$D=(-11)^2-4(1)(10)=81>0$$

$$x=\dfrac{11\pm\sqrt{81}}{2(1)}=\dfrac{11\pm9}{2}$$

$$x_1=\dfrac{11-9}{2}=1, x_2=\dfrac{11+9}{2}=10$$

$$distance=10-1=9$$

The horizontal distance from one end of the bridge to the other end is 9 m.