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Q.6

Expert-verified

Algebra 1

Found in: Page 566

Book edition Student Edition

Author(s) Carter, Cuevas, Day, Holiday, Luchin

Pages 801 pages

ISBN 9780078884801

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Short Answer

Consider $y = x^{2} - 5 x + 4$
Find the coordinates of the vertex. Is it a maximum or minimum point?

The coordinate of the vertex is $(\frac{5}{2}, - \frac{9}{4})$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Define the standard form of the quadratic function.

A quadratic function, which is written in the form, $y = a x^{2} + b x + c$ , where, $a \neq 0$ is called the standard form of the quadratic function

Step 2. Define the maximum or minimum point of the function y=ax2+bx+c.

The graph of the function $y = a x^{2} + b x + c,$

Opens upward and has a minimum value at $x = - \frac{b}{2 a}$ , when $a > 0$ .

Opens downward and has a maximum value at $x = - \frac{b}{2 a}$ , when $a < 0$ .

Step 3. Define the vertex of the function y=ax2+bx+c.

The maximum or minimum point of the function $y = a x^{2} + b x + c$ , is called the vertex.

Step 4. Calculate the vertex of the function y=x2−5x+4.

Compare the quadratic function $y = x^{2} - 5 x + 4$ with the standard equation of the quadratic function, $y = a x^{2} + b x + c$ .

$a = 1, b = - 5, c = 4$

Substitute, $a = 1$ and role="math" localid="1647752801168" $b = - 5$ in $x = - \frac{b}{2 a}$ .

$x = - \frac{(- 5)}{2 (1)} x = \frac{5}{2}$

Since,

So, the graph of the function opens upward and has a minimum point at $x = \frac{5}{2}$ .

Substitute $x = \frac{5}{2}$ in $y = x^{2} - 5 x + 4$ .

$y = {(\frac{5}{2})}^{2} - 5 (\frac{5}{2}) + 4$

$= \frac{25}{4} - \frac{25}{2} + 4 = \frac{25 - 2 (25) + 4 (4)}{4} = \frac{25 - 50 + 16}{4} = \frac{41 - 50}{4} = - \frac{9}{4}$

Hence, vertex $= (\frac{5}{2}, - \frac{9}{4})$

Therefore, the coordinate of the vertex is $(\frac{5}{2}, - \frac{9}{4})$ .

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