Remember the shortcut way for graphing quadratic equations
- A quadratic function has graph as parabola
- Hence on both sides of vertex the parabola is symmetric and axis of symmetry is vertex x values .
- y on both sides for -x and +x is same
#1
Vertex
As a is positive parabola facing upwards
Find y for same x distance from vertex
I took 3-1=2 and 3+1=4
- f(2)=2(2-3)²-1=2(-1)²-1=2-1=1
- f(4)=2(4-3)²-1=1
Now plot vertex and these two points (2,1) and (4,1) on graph then draw a parabola by freehand
#2
- y=(x-2)(x+4)
- y=x²+4x-2x-8
- y=x²+2x-8
Convert to vertex form
Vertex at (-1,-9)
Same take two equidistant x values
Let's take -1-1=-2 and -1+1 =0
- f(-2)=(-2+1)²-9=1-9=-8
- f(0)=(1)²-9=-8
Put (-1,-9),(-2,-8),(0,-8) on graph and draw a freehand parabola
#3.
Yes it can be verified by finding the coordinate theoretically on putting them on function then can be verified through putting them on graph whether they matches or not
Answer:
Try 116?
Explanation:
First, let's us know the formula for finding area of a rectangle.
Formula - A = l * w
In the image, it shows the length (12) and width (10)
Length = L
Width = W
Now, what is 12 x 10? 120
Because we need the shaded region and the little box isn't apart of it, we need to find the area of that box to and subtract it from big box
The length is 2 and the width is 2, shown by the image.
2x2 = 4
120-4 = 116
2 is the greatest common factor. Neither 4 nor 8 divides into 8-22 without a remainder.
Thus, 8 - 22 = 2(4 -11) (answer)
- 7/20 + -2/5 ? 16 / 20, I'm pretty sure
Step-by-step explanation:
On solving denominator →
(16x⁴)^¼ = [(2x)⁴]^¼ = 2x ....eq.1 { 16x⁴ = (2x)⁴ }
Numerator = 2x²
fraction = 2x² / 2x { from eq.1 }
Simplest form = x
