The graph of y = ax 2 + bx + c is a parabola that opens up and has a vertex at (-2, 5). what is the solution set of the related equation 0 = ax 2 + bx + c?

a nonlinear function that can be written on the standard form

is called a quadratic function.

all quadratic functions has a u-shaped graph called a parabola. the parent quadratic function is

the lowest or the highest point on a parabola is called the vertex. the vertex has the x-coordinate

the y-coordinate of the vertex is the maximum or minimum value of the function.

a > 0                                    parabola opens up                                      minimum value

a < 0                                      parabola opens down                          maximum value

a rule of thumb reminds us that when we have a positive symbol before  x2we get a happy expression on the graph and a negative symbol renders a sad expression.

the vertical line that passes through the vertex and divides the parabola in two is called the axis of symmetry. the axis of symmetry has the equation

the y-intercept of the equation is c.

when you want to graph a quadratic function you begin by making a table of values for some values of your function and then plot those values in a coordinate plane and draw a smooth curve through the points.

example

graph

make a table of value for some values of x. use both positive and negative values!

graph the points and draw a smooth line through the points and extend it in both directions

notice that we have a minimum point which was indicated by a positive a value (a = 1). the vertex has the coordinates (-1, 0)  which is what you will get if you use the formula for the x-coordinate of the vertex

and that the line has an y-intercept of (0, 1) which could have been determined from the c-value which is 1.

if you have an absolute value of a that is greater than 1 the parabola will be narrower than the parental quadratic function. and the opposite that if you have a absolute value of a that is less than 1 then the parabola will be wider than the parental quadratic function.

here  you can get a visual of your quadratic equations

answer: the answer is 35%

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.