Tags

, , , ,

HOW DO YOU PLOT A QUADRATIC FUNCTION OR EQUATION ON A GRAPH?

EXAMPLE I

Given that f(x)=x²-2x-8

STEP 1 – FINDING THE MAXIMUM OR MINIMUM POINT

Since a is +ve number, then the graph must be U in shape and has minimum point. Finding the minimum point by using either one of the 2 known methods

METHOD I

METHOD II

f(x)=x²-2x-8

x²-2x +(-1)²- 8 – (-1)²

x²-2x +(-1)²- 8 – 1

(x-1)²- 9

f(x)=x²-2x-8from the function given above;a=1; b=-2; c=-8to find axis symmetry, applies formula  –b/2a = -(-2)/2(1) = 1

when x=1, apply back into the function to find f(x) in order to find the min point; f(x)=1² – 2(1) – 8 = -9

Min point is (1,-9) where the axis symmetry  is when x=1 Min point is (1,-9) where the axis symmetry is when x=1

STEP 2 – CHECKING ON THE TYPE OF ROOTS FOR THE FUNCTION/EQUATION

b²-4ac = (-2)²-4(1)(-8) = 4+32=36

Since 36 > 0, the function/equation has 2 real roots and the function/equation will intercepts with x-axis on 2 different coordinates

STEP 3 – FINDING THE ROOTS OR THE COORDINATES THAT INTERCEPT WITH X-AXIS

When the function/equation intercepts with x-axis, the f(x) or y = 0

x²-2x-8=0

(x – 4)(x + 2) = 0 ; x = 4 or x = -2 and the coordinates on the x-axis shall be (4,0) &  (-2,0)

STEP 4 – FINDING THE COORDINATE THAT INTERCEPT WITH Y-AXIS

When the function/equation intercepts with y-axis, x = 0

Therefore: f(x) or y = 0² – 2(0) – 8 = -8 and the coordinate shall be (0, -8)

STEP 5 – PLOT THE GRAPH

Upon obtaining all the relevant points and coordinates and subject to the nature of the function/equation (which in this case shall be U), plotting the graph shall be easier and accurate.