Example: f(x) = 2(x-3)²+4
It tells us that the Graph of f(x) has Vertex Coordinates (3,4). The Vertex (3,4) is a result of shifting the Standard Parabola f(x) = ax² with its Vertex (0,0) 3 units right and 4 units up. See image below.
Since the leading coefficient a=2 is greater than 0 the parabola opens to the top which implies that this function’s Vertex (3,4) is a minimum.
2) f(x) = ax²+bx+c is called Standard Form.
It tells us that the Graph of f(x) has a Y-intercept at (0,c) .
Example: f(x) = 2x²+4x+5 . Its graph has Y-Intercept 5 as shown in image below.
3) f(x) = a(x-r)(x-s) is called the Factored Form.
It tells us that the Graph of f(x) has the 2 Zeros x=r , x=s .
Example: f(x)=2(x-3)(x-5) its graph has the 2 Zeros: x=3 and x=5 as shown in graph below.
Given the Standard Form of a Quadratic Equation f(x)=ax²+bx+c there is a quick and a longer way called “Complete The Square Method” to find the Vertex Form:
1) The quick way to find the Vertex Coordinates (h,k) uses the formula
h = -b/(2*a) and k = f(h)
Once computed, (h,k) along with the leading coefficient a are plugged into f(x) = a(x-h)²+k .
Example: Given the Standard Form f(x)=1x²+8x+3 then
h= -b/(2*a) = -8/(2*1) = -4 and thus
k = f(-4) = (-4)²+8(-4)+3 = 16-32+3 = -13
Replacing (-4,-13) along with a=1 yields the Vertex Form: f(x) = 1(x+4)²-13 .
See the image below: It shows the Vertex (-4,-13) of the Function f(x)=1x²+8x+3 .
2) The Complete-The-Square Method finds the Vertex Coordinates (h,k) by converting the Standard Form f(x)=ax²+bx+c into Vertex Form f(x) = a(x-h)²+k in 3 Steps.
Example: We redo the above example using the Complete the Square Method.
Step 1: Subtract 3 from y=x²+8x+3 to get y-3=x²+8x .
Step 2: To Complete the Square on x²+8x we must take half of 8 which is 4 and compute
(x+4)² = x²+8x+16 .
Our function contains x²+8x. Since 16 is missing we simply add 16 to both sides:
y-3 + 16=x²+8x+16
We just Completed The Square! We can now simplify:
y+13=(x+4)²
Step 3: Subtracting 13 yields the Vertex Form y = (x+4)²-13 .
Comparing this to the general Vertex Form y = a(x-h)²+k we can conclude that the given Quadratic Equation has the Vertex (h,k) = (-4,-13) .
The two above methods work just fine: We are given the Standard Form
y = -3x² -6x-2 .
Using the fast method, we first compute the x-coordinate of the Vertex
h= -b/(2a) = -(-6)/(2*-3) = 6/-6 = -1 .
Next, compute the y-coordinate of the Vertex by plugging h=-1 into the given equation:
k= -3*(-1)²-6*(-1)-2 = -3+6-2 = 1 .
Therefore, the Vertex is
(h,k)=(-1,1) .
Thus, the Vertex form of the Parabola is y=-3*(x+1)²+1
Easy, wasn’t it?
Tip: When using our Vertex Form Calculator to solve -3x²-6x-2=0 we must enter the 3 coefficients as
a=-3, b=-6 and c=-2.
Then, the calculator will find the vertex (h,k)=(-1,1) Step by Step.
Get it now? Try the Vertex Form Calculator below a few more times.
y=a(x-h)²+kwhere (h,k) are the Vertex Coordinates.
y=ax²+bx+c
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