Write the equation of the parabola graphed below given the y-intercept is (0, 1) Brainliest gets 45 points so please help ASAP

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9.5 Graphing Parabolas

LEARNING OBJECTIVES

Graph a parabola.

Find the intercepts and vertex of a parabola.

Find the vertex of a parabola by completing the square.

The Graph of a Quadratic Equation

We know that any linear equation with two variables can be written in the form

y

=

m

x

+

b

and that its graph is a line. In this section, we will see that any quadratic equation of the form

y

=

a

x

2

+

b

x

+

c

has a curved graph called a parabola.

Two points determine any line. However, since a parabola is curved, we should find more than two points. In this text, we will determine at least five points as a means to produce an acceptable sketch. To begin, we graph our first parabola by plotting points. Given a quadratic equation of the form

y

=

a

x

2

+

b

x

+

c

, x is the independent variable and y is the dependent variable. Choose some values for x and then determine the corresponding y-values. Then plot the points and sketch the graph.

y =  x² - 2x + 1

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

From the graph (h, k) = (2, - 1), thus

y = a(x - 2)² - 1

To find a substitute a point on the curve (0, 1) into the equation

1 = 4a - 1 ( add 1 to both sides )

4a = 2 ( divide both sides by 4 )

a =  = , hence

y = (x - 2)² - 1 ← in vertex form

Expand the factor and simplify

y = (x² - 4x + 4) - 1

 =  x² - 2x + 2 - 1

 =  x² - 2x + 1 ← in standard form

answer: idk

step-by-step explanation: