Tags
axis symmetry, finding roots of equation, max or min point, plotting a graph using quadratic equation or function, type of roots
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.
vaishu said:
10q so much..it very helpful for me..
Hajar Radhiah Mohd Som said:
sir…how i want to solve this problem..
-the minimum value of the quadratic equation f(x)=x²-6hx+10h²+4 is a²+4h,where a and h are constant.
a.) show that a=h-2 by using the method of completing square
b.) hence,find the values of h and a if the function has an axis of symmetry x= a²-4
unclezul said:
Dear Hajar,
First of all you must be able to master the Completing the Square method, which Uncle quite confident that you could do just that.
x²-6hx+10h²+4
x²-6h+(-3h)²+10h²+4-(-3h)²
(x-3h)² +10h²+4-9h²
(x-3h)² +10h²-9h²+4
(x-3h)² +h²+4
From here, we can obtain the min point at (3h,h²+4)
where axis of symmetry is when x = 3h with a minimum value of h²+4
(a) Since the min value is given as a²+4h, then
a²+4h = h²+4
a² = h²-4h+4 = (h-2)²
a = h-2
(b) x= a²-4 = 3h where a = h-2;
(h-2)²-4 = 3h
h²-4h+4-4 = 3h
h²-4h-3h=0
h²-7h=0
h(h-7)=0
h=0 or h=7
When h=0, a=0-2 = -2
When h=7, a=7-2 = 5
nadzirah safuraa said:
sir , how i want to slve this problem :
given y² = x² – 2qx has a minimum value of 6
without using differentiation method , find the possible values of q
Afiq Zukri said:
Sir,i want to ask some question,why there is 0 in
f(x) or y = (this ->0²(0)<-) – 8 = -8 —-step 4
unclezul said:
Dear Afiq,
Whenever the graph touches the y-axis, x will always be 0. The same principle applies when the graph touches x-axis, then y will always be 0