Question

Maths functions question!!

Answer 1

Answer:

5)  DE = 7 units and DF = 4 units

6)  ST = 8 units

$\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units$

8)  x ≤ -3 and x ≥ 3

Step-by-step explanation:

Information from Parts 1-4:

brainly.com/question/28193969

• $f(x)=-x+3$

• $g(x)=x^2-9$

• A = (3, 0)  and C = (-3, 0)

Part (5)

Points A and D are the points of intersection of the two functions.  

To find the x-values of the points of intersection, equate the two functions and solve for x:

$\implies g(x)=f(x)$

$\implies x^2-9=-x+3$

$\implies x^2+x-12=0$

$\implies x^2+4x-3x-12=0$

$\implies x(x+4)-3(x+4)=0$

$\implies (x-3)(x+4)=0$

Apply the zero-product property:

$\implies (x-3)= \implies x=3$

$\implies (x+4)=0 \implies x=-4$

From inspection of the graph, we can see that the x-value of point D is negative, therefore the x-value of point D is x = -4.

To find the y-value of point D, substitute the found value of x into one of the functions:

$\implies f(-4)=-(-4)=7$

Therefore, D = (-4, 7).

The length of DE is the difference between the y-value of D and the x-axis:

⇒ DE = 7 units

The length of DF is the difference between the x-value of D and the x-axis:

⇒ DF = 4 units

Part (6)

To find point S, substitute the x-value of point T into function g(x):

$\implies g(4)=(4)^2-9=7$

Therefore, S = (4, 7).

The length ST is the difference between the y-values of points S and T:

$\implies ST=y_S-y_T=7-(-1)=8$

Therefore, ST = 8 units.

Part (7)

The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x.  To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:

$\implies f(x)-g(x)=QR$

$\implies -x+3-(x^2-9)=\dfrac{45}{4}$

$\implies -x+3-x^2+9=\dfrac{45}{4}$

$\implies -x^2-x+\dfrac{3}{4}=0$

$\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)$

$\implies 4x^2+4x-3=0$

$\implies 4x^2+6x-2x-3=0$

$\implies 2x(2x+3)-1(2x+3)=0$

$\implies (2x-1)(2x+3)=0$

Apply the zero-product property:

$\implies (2x-1)=0 \implies x=\dfrac{1}{2}$

$\implies (2x+3)=0 \implies x=-\dfrac{3}{2}$

As the x-value of points M, Q and P is negative, x = -³/₂.

Length OM is the difference between the x-values of points M and the origin O:

$\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}$

Therefore, OM = ³/₂ units.

Part (8)

The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.

Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.

Answer 2

So we know that these are the 2 lines
y = -x + 3
y = x^2 -9

They intersect at D so we can put them equal to each other
x^2 -9 = -x +3
And solve by moving everything to one side:
x^2 +x -12 = 0
Then solve the quadratic by factorisation:
(x+4)(x-3)=0
So x = -4 or 3
Which we can sub into
y = -x + 3
To get
y = 4+3 = 7
(-4,7)
And
y = -3 + 3 = 0
(3,0)

We already know (3,0) is where it crosses the x axis so the coordinates of d must be (-4,7)
DE is the distance from the x axis which is just the y value = 7
DF is the distance from the y axis which is the x value except positive because it is a length = 4

I assume d f and s make a straight line so the coordinates of s are (4,7) since the curve is symmetrical in the y axis

The distance between (4,-1) and (4,7) is 7- -1 = 8

The next one’s asking what the x coordinate is for q/m/r (they’re the same) if q’s y coordinate - r’s y coordinate = 45/4
so:
(-x + 3) - (x^2 -9) = 45/4
-x^2 -x + 12 = 45/4
Which can be solved first by multiplying out the fraction and rearranging
-4x^2 -4x + 48 = 45
0 = 4x^2 + 4x -3
Then by factorisation
4x^2 +6x - 2x - 3
(2x-1)(2x+3)
So x = 1/2 or -1.5
We know in this case x is negative as it is left of the y axis so OM = 1.5

The next question is asking when the curve goes above the x axis
Since we know where it crosses the x axis, we know that the answer is
x less than or equal to -3
And
x greater than or equal to 3