Need help with two questions.

Ok, ive been set these two questions to do, and i cant remember at all how to do them.

So if someone could just go through it with me i would appreciate it loads.

1. f(x) = x2 – kx + 9, where k is a constant.

(a) Find the set of values of k for which the equation f(x) = 0 has no real solutions.

Given that k = 4,

(b) express f(x) in the form (x – p)2 + q, where p and q are constants to be found,

(c) write down the minimum value of f(x) and the value of x for which this occurs.

2.
f(x) = 9 – (x – 2)2

(a) Write down the maximum value of f(x).

(b) Sketch the graph of y = f(x), showing the coordinates of the points at which the graph meets the coordinate axes.

So if anyone could just talk me through this and show me what im supposed to be doing i would actually love you!

Reply 1

Glutamic Acid

1a) You need to use the discriminant, which is

b^2-4ac

. When this is equal to zero, there is one repeated root. When it is more than 0 there are 2 distinct real roots, when it is less than 0 there are no real roots.

the values a, b and c are taken when the equation is in the form

ax^2+bx+c = 0

. So we have a = 1, b = -k and c = 9
So

b^2-4ac < 0\Rightarrow k^2-36 < 0

Then just solve like a normal inequality.

Reply 2

Glutamic Acid

b) This is the technique called completing the square. To do this we half the value of b when the equation is in the form

ax^2+bx+c=0

and put it in brackets, which are then squared. This gives

(x-2)^2+q = x^2-4x+9=0

So which value of q gives 9 as the constant?

Reply 3

Glutamic Acid

c) The least value of a function is also the vertex and it occurs when a in ax^2 is positive. Say we have the equation

(x-p)^2+q = 0

. The smallest value will occur when the brackets are equal to 0 (all other values are positive since the brackets are squared). So when x = p, the least value will be q.

Reply 4

Dadeyemi

2)a) I'm assuming you mean

9-(x-2)^2

well this is just remembering the stationary point of

(x-a)^2+b

is at the point(a,b) apply this to the

9-(x-2)^2

and you have the answer.
b)multiply it out to get

y=-x^2-4x-5

mark on the maximum point, which you just found, and the points where it crosses the y and a axis.
for the y cross just check the last coefficient (-5) and for the x-axis crossing points just solve f(x)=0
this gets (-x+1)(x+5)=0 i think, so x = 1 and -5 when y =0

draw these points and use them to draw the graph