Find the vertex:
Determine if the parabola will open up or down:
Since the value of a is positive, the parabola will open upwards
Find the x and y intercepts:
Factor 
to find the x intercepts
(x+8)(x+2)
x + 8 = 0
x = -8
x + 2 = 0
x = -2
x intercepts:
x = -2
x = -8
Find the y intercept
Hence, C. Graph A best represents the quadratic function
Answer:
Any points in the shaded region including (2,-2) and (-3,-8)
Step-by-step explanation:
Convert the line into slope intercept form and graph it.
2x-y > 1 becomes -y>1-2x. Divide both sides by -1 and you get y<2x-1. Graph it with the shaded area on the right and a dashed line.
Any point which falls within the shaded red of the graph is a solution. No points on the line since it is not equal to (its dashed) are solutions. Check the location of your points to verify that they fall within this area.
(-3, -8) ---Yes
(-1, -3) ---No
(0, 5) --- No
(1, 6) --- No
(2, -2) ---Yes
<h3>Refer to the diagram below</h3>
- Draw one smaller circle inside another larger circle. Make sure the circle's edges do not touch in any way. Based on this diagram, you can see that any tangent of the smaller circle cannot possibly intersect the larger circle at exactly one location (hence that inner circle tangent cannot be a tangent to the larger circle). So that's why there are no common tangents in this situation.
- Start with the drawing made in problem 1. Move the smaller circle so that it's now touching the larger circle at exactly one point. Make sure the smaller circle is completely inside the larger one. They both share a common point of tangency and therefore share a common single tangent line.
- Start with the drawing made for problem 2. Move the smaller circle so that it's partially outside the larger circle. This will allow for two different common tangents to form.
- Start with the drawing made for problem 3. Move the smaller circle so that it's completely outside the larger circle, but have the circles touch at exactly one point. This will allow for an internal common tangent plus two extra external common tangents.
- Pull the two circles completely apart. Make sure they don't touch at all. This will allow us to have four different common tangents. Two of those tangents are internal, while the others are external. An internal tangent cuts through the line that directly connects the centers of the circles.
Refer to the diagram below for examples of what I mean.
X^4 + x^3 + x^2 + x = 0
x(x^3 + x^2 + x + 1) = 0
x[x^2(x + 1) + 1(x + 1)] = 0
x(x + 1)(x^2 + 1) = 0 ⇔ x = 0 or x + 1 = 0 or x^2 + 1 = 0
x = 0 or x = -1 or x^2 = -1 -false
Answer: x= 0 or x = -1.
Answer:
5
Step-by-step explanation:
→ Let x = √20 + √20 .....
x
→ Square both sides
x² = 20 + √20 +√20 + √20
→ Replace √20 +√20 + √20 with x
x² = 20 + x
→ Move everything to the left hand side
x² - 20 - x = 0
→ Factorise
( x + 4 ) ( x - 5 )
→ Solve
x = -4 , 5
→ Discard negative result
x = 5
