Answer Vertex : (1,2) Axis of Symmetry : x = 1 y-intercept : 1 Explanation Given Information: Given equation is : y = − x 2 + 2 x + 1 Concept Used : Quadratic Functions: Quadratic functions are non-linear and are of the form: f ( x ) = a x 2 + b x + c ; a ≠ 0 This is the standard form of Quadratic function. Shape of the graph of a Quadratic function is called Parabola. Parabolas are symmetric about a central line called Axis Of Symmetry. The axis of symmetry intersects a parabola only at one point , called the Vertex. Parent Function: y = x 2 Standard Form : f ( x ) = a x 2 + b x + c ; a ≠ 0 Type of Graph: Parabola Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves. x = − b 2 a y-intercept = The y coordinate of the point at which the graph intersects y-axis = c Graph: If a > 0 , The graph of a x 2 + b x + c opens upward. The lowest point on the graph is the minimum. If a < 0 , The graph of a x 2 + b x + c opens downward. The highest point on the graph is the maximum. The maximum or the minimum is the vertex. Calculation: In the equation y = − x 2 + 2 x + 1 ; comparing it with standard form of quadratic equation , we have a = -1 , b = 2 , c = 1 . Find Axis of Symmetry : y = a x 2 + b x + c ........................... [ s tan d a r d _ f o r m ] x = − b 2 a ..................................... [ f o r m u l a _ o f _ a x i s _ o f _ s y m m e t r y ] x = − 2 2 ( − 1 ) = 1.......................... [ a = − 1 , b = 2 ] The equation of axis of symmetry is x = 1 Find the Vertex : Since axis of symmetry passes through the vertex , we have x coordinate of the vertex (x,y) is x=1 Substituting the value of x =1 in the equation y = − x 2 + 2 x + 1 y = − x 2 + 2 x + 1 y = − ( 1 ) 2 + 2 ( 1 ) + 1..................... [ x = 1 ] y = − 1 + 2 + 1................................ [ s i m p l i f y ] y = 2 The vertex is at (1,2) Find y-intercept : The y-coordinate of the point at which the graph cuts the y-axis is the y-intercept. Hence , substituting x=0 in the equation y = − x 2 + 2 x + 1 , we will get the y-intercept. y = − x 2 + 2 x + 1 y = − ( 0 ) 2 + 2 ( 0 ) + 1..................... [ x = 0 ] y = 1 Or y-intercept always occurs at (0,c) and c = 1 here. So, y-intercept = 1 and is located at (0,1)

1st Edition

ISBN: 9780078884801

Author: McGraw-Hill/Glencoe

Publisher: Glencoe/McGraw-Hill School Pub Co

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Question

Chapter 9.1, Problem 10CYU

To determine

To Find:

The vertex , axis of symmetry and y − intercept.

Expert Solution & Answer

Answer to Problem 10CYU

Vertex : (1,2)

Axis of Symmetry : x = 1

y-intercept : 1

Explanation of Solution

Given Information:

Given equation is :

$y=−x2+2x+1$

Concept Used :

Quadratic Functions:
Quadratic functions are non-linear and are of the form:

$f(x)=ax2+bx+c;a≠0$

This is the standard form of Quadratic function.

Shape of the graph of a Quadratic function is called Parabola.
Parabolas are symmetric about a central line called Axis Of Symmetry.
The axis of symmetry intersects a parabola only at one point , called the Vertex.
Parent Function:

$y=x2$

Standard Form :

$f(x)=ax2+bx+c;a≠0$

Type of Graph:

Parabola

Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

$x=−b2a$

y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
Graph:

Algebra 1, Chapter 9.1, Problem 10CYU

If a > 0 ,
The graph of $ax2+bx+c$ opens upward.
The lowest point on the graph is the minimum.
If a < 0 ,
The graph of $ax2+bx+c$ opens downward.
The highest point on the graph is the maximum.
The maximum or the minimum is the vertex.

Calculation:

In the equation $y=−x2+2x+1$ ; comparing it with standard form of quadratic equation , we have a = -1 , b = 2 , c = 1 .
Find Axis of Symmetry :

$y=ax2+bx+c...........................[standard_form]x=−b2a.....................................[formula_of_axis_of_symmetry]x=−22(−1)=1..........................[a=−1,b=2]$

The equation of axis of symmetry is x = 1

Find the Vertex :

Since axis of symmetry passes through the vertex , we have x coordinate of the vertex (x,y) is x=1

Substituting the value of x =1 in the equation $y=−x2+2x+1$

$y=−x2+2x+1 y=− (1) 2 +2(1)+1.....................[x=1]y=−1+2+1................................[simplify]y=2$

The vertex is at (1,2)

Find y-intercept :

The y-coordinate of the point at which the graph cuts the y-axis is the y-intercept.

Hence , substituting x=0 in the equation $y=−x2+2x+1$ , we will get the y-intercept.

$y=−x2+2x+1 y=− (0) 2 +2(0)+1.....................[x=0]y=1$

y-intercept always occurs at (0,c) and c = 1 here.

So, y-intercept = 1 and is located at (0,1)

Chapter 9.1, Problem 9CYU

Chapter 9.1, Problem 11CYU

Chapter 9 Solutions

Algebra 1

Ch. 9.1 - Prob. 67PPS Ch. 9.1 - Prob. 68HP Ch. 9.1 - Prob. 69HP Ch. 9.1 - Prob. 70HP Ch. 9.1 - Prob. 71HP Ch. 9.1 - Prob. 72HP Ch. 9.1 - Prob. 73HP Ch. 9.1 - Prob. 74HP Ch. 9.1 - Prob. 75STP Ch. 9.1 - Prob. 76STP

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The vertex , axis of symmetry and y − intercept.