Question

Find the answer for given questions? Slope- intercept? Point slope standard form?

Answer 1

The equation in slope-intercept form will be;

`y=-(1)/(2)x+1`

The function in point-slope form is;

`y-2=-(1)/(2)(x+2)_{}`

The standard form of the equation is;

`x+2y=2`

Step-by-step explanation:

Given the function f(x);

`\begin{gathered} f(-2)=2 \\ f(8)=-3 \end{gathered}`

Firstly, let us find the slope;

`\begin{gathered} m=(f(8)-f(-2))/(8-(-2))=(-3-2)/(8+2)=(-5)/(10) \\ m=-(1)/(2) \end{gathered}`

we can then solve for the constant term;

`\begin{gathered} y=mx+b \\ at\text{ f(-2)=2;} \\ 2=-(1)/(2)(-2)+b \\ 2=1+b \\ b=2-1 \\ b=1 \end{gathered}`

The equation in slope-intercept form will be;

`y=-(1)/(2)x+1`

Applying the point-slope form of equation;

`y-y_2=m(x-x_2)`

Substituting the second point;

`\begin{gathered} y-2=-(1)/(2)(x-(-2)) \\ y-2=-(1)/(2)(x+2)_{} \end{gathered}`

The function in point-slope form is;

`y-2=-(1)/(2)(x+2)_{}`

The standard form of the equation can be written as;

`\begin{gathered} y=-(1)/(2)x+1 \\ (1)/(2)x+y=1 \\ \text{multiply through by 2} \\ x+2y=2 \end{gathered}`

The standard form of the equation is;

`x+2y=2`