Help needed again on c1

f(x) = x³ - x² - 7x + c, where c is a constant

Given that f(4) = 0

a) find the value of c
b) factorise f(x) as the product of a linear factor and a quadratic factor.
c) Hence, show that apart from x = 4, there are no real values of x for which f(x) = 0

another question :redface:

The straight line I, has equation 4y + x = o
The straight line I₂ has equation y = 2x - 3

1) On the same axes, sketch the graphs of I₁ and I₂ , show clearly the coordinates of all points at which the graphs meet the coordinate axes.

another another one :eek:

Given that f(x) = 15 - 7x - 2x²

a) find the coordinates of all points at which the graph of Y=f(x) crosses the coordinate axes.

b) sketch the graph of y=f(x)

thanx alot :wink:

Reply 1

username9303

hey, i don't mean to be rude but if you're doing A-level maths and you can't do quadratic equations then surely theres something wrong.

a) c = -20
b) (x-4)(x²+3x+5) i think
c) just try and solve that quadratic and you will get a minus number in the square root.

can't be bothered to do the rest.

Reply 2

Geminus

The straight line I, has equation 4y + x = o
The straight line I2 has equation y = 2x - 3

1) On the same axes, sketch the graphs of I1 and I2 , show clearly the coordinates of all points at which the graphs meet the coordinate axes.

Think back to GCSE maths. If you have the equation of a line in the form y=mx + c, what does the value of 'm' and 'c' tell you about the graph? Post again if you're not sure.

Given that f(x) = 15 - 7x - 2x²

a) find the coordinates of all points at which the graph of Y=f(x) crosses the coordinate axes.

b) sketch the graph of y=f(x)

a) Think about what the values of x and y will be when the graph crosses the axes. When it crosses the x axis, y must equal 0, yes? When it crosses the y axis, x must be equal 0. Now look at the equation, plug in x=0, and then solve the quadratic to find x when y=0.

b) Using the roots of the graph which you found in part a, you'll be able to sketch the graph. Remember that the value of your x^2 term is negative; consider what affect this will have on the shape of the graph.

Reply 3

Malik

Milli

f(4)=0 Means when x=4 f(x)=0
4³ - 4² - 7(4) + c=0
64-16-28+c=0
20+c=0
c=-20

(b) f(x)=x³ - x² - 7x -20
(x-4)(x²+3x+5)=0
No real values for x²+3x+5 so only real value x=4

2)
4y+x=0
Subsitute x=-4y into y=2x-3
y=2(-4y)-3
y=-8y-3
y=-1/3
Substitute y=-1/3 back into x=-4y
x=-4(-1/3)
x= 4/3
They meet at (4/3,-1/3)

To find where they meet at axes subsitute x=0 y=0
when x=0
4y+0=0
y=0
(0,0)

y=2x-3 When x=0
y=-3
(0,-3)

Just draw a rough sketch of these lines all you need to do for marks is label all those coordinates.

3) f(x)=15-7x-2x²
They just want you to solve it so:
-2x²-7x+15=0
2x²+7x-15=0
(x+5)(2x-3)
x=-5 x=3/2
(-5,0)(3/2,0)
Just plot those points and label them

Reply 4

Mechs

Milli

f(x) = x³ - x² - 7x + c, where c is a constant

a) sub f(4) = 4³ - 4² - 7(4) + c = 0
...............= 64 - 16 - 28 + c = 0
................= 48 - 28 = -c, -c = 20, c = -20

b) 0 = x³ - x² - 7x - 20
(x-4) is a factor, x³ - x² - 7x - 20 / (x-4)

......... (x² + 3x + 5)
(x-4) / x³ - x² - 7x - 20
.......- x³ - 4x²
............... 3x² - 7x
.............- 3x² -12x
...................... 5x - 20
.....................- 5x - 20

So (x-4)(x² + 3x + 5)

c) x² + 3x + 5 = 0
... (x - 3/2)² - 9/4 + 5 = 0
... x = 3/2 ± √-11/4

The √-11/4 shows that no real roots for x² + 3x + 5, hence no other roots except x=4 for the previous equation. (discriminant)

Given that f(x) = 15 - 7x - 2x²

a) find the coordinates of all points at which the graph of Y=f(x) crosses the coordinate axes.

f(x) = 15 - 7x - 2x²
0 = 2x² + 7x - 15
0 = x² + 7/2x - 15/2
0 = (x + 7/4)² - 49/16 - 15/2
x = -7/4 ± √(169/16)
x = 3/2, -5

f(x) crosses x-axis at -5, 3/2 and y-axis at 15.

Sorry don't do graph sketches! :redface: