How to set up the equation

Maths functions <br> please help!

Vlad [161]

Answer:

$\textsf{1)} \quad f(x)=-x+3$

2) A = (3, 0) and C = (-3, 0)

$\textsf{3)} \quad g(x)=x^2-9$

4) AC = 6 units and OB = 9 units

Step-by-step explanation:

Given functions:

$\begin{cases}f(x)=mx+c\\g(x)=ax^2+b \end{cases}$

Given points:

H = (-1, 4)
T = (4, -1)

As points H and T lie on f(x), substitute the two points into the function to create two equations:

$\textsf{Equation 1}: \quad f(-1)=m(-1)+c=4 \implies -m+c=4$

$\textsf{Equation 2}: \quad f(4)=m(4)+c=-1 \implies 4m+c=-1$

Subtract the first equation from the second to eliminate c:

$\begin{array}{r l} 4m+c & = -1\\- \quad -m+c & = \phantom{))}4\\\cline{1-2}5m \phantom{))))}}& = -5}\end{aligned}$

Therefore m = -1.

Substitute the found value of m and one of the points into the function and solve for c:

$\implies f(4)=-1(4)+c=-1$

$\implies c=-1-(-4)=3$

Therefore the equation for function f(x) is:

$f(x)=-x+3$

Function f(x) crosses the x-axis at point A. Therefore, f(x) = 0 at point A.

To find the x-value of point A, set f(x) to zero and solve for x:

$\implies f(x)=0$

$\implies -x+3=0$

$\implies x=3$

Therefore, A = (3, 0).

As g(x) = ax² + b, its axis of symmetry is x = 0.

A parabola's axis of symmetry is the midpoint of its x-intercepts.

Therefore, if A = (3, 0) then C = (-3, 0).

Points on function g(x):

A = (3, 0)
G = (1, -8)

Substitute the points into the given function g(x) to create two equations:

$\textsf{Equation 1}: \quad g(3)=a(3)^2+b=0 \implies 9a+b=0$

$\textsf{Equation 2}: \quad g(1)=a(1)^2+b=-8 \implies a+b=-8$

Subtract the second equation from the first to eliminate b:

$\begin{array}{r l} 9a+b & = \phantom{))}0\\- \quad a+b & =-8\\\cline{1-2}8a \phantom{))))}}& = \phantom{))}8}\end{aligned}$

Therefore a = 1.

Substitute the found value of a and one of the points into the function and solve for b:

$\implies g(3)=1(3^2)+b=0$

$\implies 9+b=0\implies b=-9$

Therefore the equation for function g(x) is:

$g(x)=x^2-9$

The length AC is the difference between the x-values of points A and C.

$\implies x_A-x_C=3-(-3)=6$

Point B is the y-intercept of g(x), so when x = 0:

$\implies g(0)=(0)^2-9=-9$

Therefore, B = (0, -9).

The length OB is the difference between the y-values of the origin and point B.

$\implies y_O-y_B=0-(-9)=9$

Therefore, AC = 6 units and OB = 9 units

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