# Write the slope intercept form of the equation of the line which goes through the point with coordinates (3,1) and the vertex of the parabola y=x^2-4x+7.

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You should remember the slope intercept form of equation of line such that:

`y = mx + b`

m represents the slope

b represents y intercept

The problem provides two points, one is given,(3,1) and the other point needs to be evaluated since it is the vertex of parabola y = `x^2 - 4x + 7.`

You should remember the formulas that give the coordinates of vertices of parabola such that:

`x = -b/(2a) , y = (4ac - b^2)/(4a)`

a,b,c represent he coefficients of quadratic equation `ax^2 + bx + c = 0` .

Identifying a,b,c yields:

`a = 1 , b = -4, c = 7`

`x = 4/2 = 2`

`y = (28 - 16)/4 => y = 12/4 = 3`

Hence, the coordinates of vertex are `(2,3).`

You may write the equation of the line that passes through the points (3,1) and (2,3) such that:

`y - 1 = (3-1)/(2-3)(x - 3) => y - 1 = -2(x - 3)`

Converting the point slope form of equation into slope intercept form yields:

`y = -2x + 7`

**Hence, evaluating the slope intercept form of equation of the line that passes through the given points yields `y = -2x + 7` .**

`y=x^2-4x+7 ` to find x use the fromula -2/2a

`a=1 b=-4 c=7` ` -(-4) = 4 ` `4/(2x1) ` `x=2`

then plug in the x into the equation.

`y=2^2-4(2)+7` ` y=4-8+7` `=3 ` `y=3 ` so the vertex is `(2,3)`

`y-1= 2/-1 (x-3)`

`y-1= -2 (x-3) ` distribute the -2

`y-1 = -2x+6 ` add 1 to both side from the y-1

`y= -2x+7`

the ans is **`y= -2x+7` **