Question

7.Given the maximum value of the quadratic function f(x)=-2x^2-5x+p is 1/2,find the value of p.pls check whether my answer and working is right. @Directrix @Jhannybean @freckles @ganeshie8

Answer 1

\[-2x^2-5x+p \\ =-2(x^2+\frac{5}{2}x)+p \\ =-2(x^2+\frac{5}{2}x+(\frac{5}{2 \cdot 2})^2)+p+2(\frac{5}{2 \cdot 2})^2 \\ =-2(x+\frac{5}{2 \cdot 2})^2+p+2( \frac{5}{2 \cdot 2})^2\]

Answer 2

added in ((5/2)/2 )^2 to complete the square

Answer 3

but it was being multiply by -2
so -2(5/4)^2+2(5/4)^2 gives 0
so I actually added in zero

Answer 4

\[=-2(x+\frac{5}{4})^2+p+2 \frac{25}{16} \\-2(x+\frac{5}{4})^2+p+\frac{25}{8}\]

Answer 5

in general,
\[y=ax^2+bx+c \\ y=a(x^2+\frac{b}{a}x)+c \\ y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ y=a(x+\frac{b}{2a})^2+c-\frac{ab^2}{4a^2} \\ y=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}\]
\[\frac{1}{2}=c-\frac{b^2}{4a} \\ \frac{1}{2}=p-\frac{(-5)^2}{4(-2)} \\ \frac{1}{2}=p-\frac{25}{-8} \\ \frac{1}{2}=p+\frac{25}{8}\]
\[p=\frac{1}{2}-\frac{25}{8} \\ p=\frac{4}{8}-\frac{25}{8}\]

Answer 6

4-25 doesn't give -29

Answer 7

hmmm 4-25=-21
I don't think -21 and 8 have any common factors

Answer 8

do you agree the max value is p+25/8?

Answer 9

and also 1/2

Answer 10

\[f(x)=-\left[ 2x^2+5x-p \right]\]
\[=-\left[ (2x+\frac{ 5 }{ 2 })^2-\frac{ 25 }{ 4 }-p \right]\]
\[=-2(x+\frac{ 5 }{ 4 })^2+\frac{ 25 }{ 4 }+p\]
\[\frac{ 25 }{ 4 }+p=-\frac{ 1 }{ 2 }\]
\[p =-\frac{ 25 }{ 4 }-\frac{ 1 }{ 2 }\]
\[p=-\frac{ 27 }{ 4 }\]

Answer 11

@freckles

Answer 12

Why do u need to multiply it by 2?

Answer 13

@freckles

Answer 14

@ganeshie8

Answer 15

\[f(x)=-2x^2-5x+p \]

factor out -2 from the first two terms only, don't touch the constant \(p\). you get

\[f(x)=-2\left(x^2+\frac{5}{2}x\right)+p \]

yes ?

Answer 16

which is same as

\[f(x)=-2\left(x^2+2\cdot \frac{5}{4}x\right)+p \]

Answer 17

\(\bf\huge\color{#ff0000}{T}\color{#ff2000}{h}\color{#ff4000}{a}\color{#ff5f00}{n}\color{#ff7f00}{k}~\color{#ffaa00}{Y}\color{#ffd400}{o}\color{#bfff00}{u}\color{#4600ff}{!}\color{#6800ff}{!}\color{#8b00ff}{!}\) @ganeshie8

Answer 18

and @freckles