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