Hello there first time poster. I'm currently on my second lesson of math tuition and have been set some questions. I'm not here to cheat I have had a crack at solving them. I would appreciate if people could confirm whether the answers are right and if there's a better method of getting the answer. Thankyou

Question 1.Now Complete

$\displaystyle x^2 + 2x - 63$

Factorise

$\displaystyle x^2 = (0+7) (0-9)$

$\displaystyle x = -7$

$\displaystyle x = 9$[these are the x crossing points]

Y crossing point

$\displaystyle y= x^2 + 2x - 63$

$\displaystyle 0^2 + 2.0 - 63$

$\displaystyle y = -63$

minimum point of x (Diff)

$\displaystyle x^2 + 2x$

$\displaystyle x^1.2 + 2x^1.1$

$\displaystyle 2x + 2x$

$\displaystyle 2x -2 = 0$

$\displaystyle -2 = 2x$

$\displaystyle x = -1$

y's mimum point putting back into equation

$\displaystyle y = x^2 + 2x -63$

$\displaystyle -1^2 + 2 x -1 -63$

$\displaystyle 1+ -2 -63$

$\displaystyle y = -60$ minumum point

Uh oh I went wrong above

It should be:$\displaystyle y=x^2=2x-63$

$\displaystyle y(-1)=(-1)^2+2.(-1)-63 = 1-2-63=-64$[thankyou earboth]

Sketch:You calculate actually the y-coordinate of the vertex of the parabola which is indeed the absolute minimum of all possible y-values.

As you may have noticed the vertex is below the y-intercept (0, -60).

Question 2.currently working on

$\displaystyle x^2-8x+16$

Factorise

$\displaystyle X^2= (0+4) (0-4)$

$\displaystyle x = -4$

$\displaystyle x = 4$[these are the x crossing points]

Y crossing point

$\displaystyle y= 0^2-6 =16$

$\displaystyle y=0^2-8.0+16=16$

16 is y

minimum point of x (Diff)

$\displaystyle x^2-8x$

$\displaystyle x^1.2-8x^1.1$

$\displaystyle 2x-10x$

$\displaystyle 2x-10=0$

$\displaystyle x=-5$

y's mimum point putting back into equation

$\displaystyle X^2-8x+16$

$\displaystyle -5-8.-5+16$

[y's minimum point is 51]had assistance of calculator for this bit

Question 3.not complete

$\displaystyle x^2-3x-4$

Factorise

$\displaystyle X^2= (0+) (0+4)$

$\displaystyle x = -1$

$\displaystyle x = -4$[these are the x crossing points...I think]

Y crossing point

$\displaystyle y= 0^2+5.0+4=4$

$\displaystyle y=x^2-3x-4$

$\displaystyle 02-3.0=-4$

minimum point of x (Diff)

$\displaystyle x^2-3x$

$\displaystyle x^1.2-3x^1.1$

$\displaystyle 2x-3x$

$\displaystyle 2x-3x=0$

$\displaystyle -3=2x$

$\displaystylex=-3/2$

y's mimum point putting back into equation

$\displaystyle X^2-3x-4$

i have changed this into a decimal point, $\displaystyle -2.25 - -4.5 = -1.75$

[y's minimum point is -1.75]I believe i have got this wrong because i have not written $\displaystyle -3/2^2 $or $\displaystyle -2.25^2$So -2.25^2 is 5.0625 - -4.5 - 4= 5.5625 (is this y?!)

I have redone the orginal question with the correction so other users can see where I went wrong and maybe learn from it .

thankyou for you time , burgoyne