Rewrite the following quadratic function in vertex form. then, determine if it has a maximum or minimum and say what that vaule is.

y = - x ^{2} + 6x + 5

Answers

Take the inverse and it is your answer. 
Srry all I can is the  give axis of symmetry (8,192)
F(x) = -3(x^2 - 8x + 16) + 7

f(x) = -3x^2 + 24x - 48 + 7

f(x) = -3x^2 + 24x - 41

f(x)=-3(x+1)^2+27

Step-by-step explanation:

Forst, simplify the expression by pulling out -3, but leave 24 out.

=-3(x^2+2x)+24

Then complete the square of the equatio inside the parenthesis (remember to subtract from the outside what you add to the inside times -3).

=-3(x^2+2x+1)+24+3

=-3(x+1)^2+27

The equation is now in vertex form.

from that, we now know that the vertex is (-1,27).

Set this equation equal to 0 to find the x intercepts:

0=-3(x+1)^2+27

-27=-3(x+1)^2

9=(x+1)^2

3 and -3=x+1

x=2 and -4

f(x) = - 2(x - \frac{1}{4})² - \frac{15}{8}

the equation of a quadratic in vertex form is

y = a(x - h)² + k

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

To obtain this form use the method of completing the square

• the coefficient of the x² term must be 1 , factor out - 2

= - 2(x² - \frac{1}{2}x) - 2

• add / subtract (half the coefficient of the x-term )² to x² - \frac{1}{2}x

= - 2(x² + 2(- \frac{1}{4})x + \frac{1}{16} - \frac{1}{16}) - 2

= - 2(x - \frac{1}{4})² + \frac{1}{8} - 2

= - 2(x - \frac{1}{4})² - \frac{15}{8}

Discussion

1. Put brackets around the first two terms

y = (-x^2 + 6x) + 5

2. Take out the common factor of -1

y = -(x^2 - 6x) + 5

3. Inside the brackets, take 1/2 of - 6 and square it

y = -(x^2 - 6x + ( - 6 / 2)^2 )  + 5

y = -(x^2 - 6x +  (- 3)^2 )  )  + 5

y = -(x^2 - 6x + 9 ) + 5

Note: Step 3 is very long. Make sure you work your way through it

4. You have added 9 inside the brackets. It is actually - 9. So add 9 outside to balance the equation out. This is the key step. Make sure you understand it.

y = - (x^2 - 6x + 9) + 5 + 9

5. Express the brackets as a square.

y = - (x + 3)^2 + 14

Discussion

The equation is now in vertex form. The minus tells you that the equation is a maximum. The maximum is located at ( - 3, 14 )

A graph follows to show the results.


Rewrite the following quadratic function in vertex form. then, determine if it has a maximum or mini

5x^2+15x-2

5(x^2+3x)-2

5(x^2+3x+2.25-2.25)-2

5(x^2+3x+2.25)-13.25

5(x+1.5)^2-13.25

final y=5(x+1.5)^2-13.25

axis of symmetry: (1.5, 13.25)

(x - 3)² = - (y - 14)

The maxima of the function is (3,14).

Step-by-step explanation:

The formula of a quadratic function is given by y = - x² + 6x + 5

We have to write this in the vertex form.

Now, y = - x² + 6x + 5

⇒ y = - (x² - 6x + 9) + 9 + 5

⇒ y - 14 = - (x - 3)²

⇒ (x - 3)² = - (y - 14)  

This is the equation of the parabola in vertex form.

This parabola has the vertex at (3,14) and the axis of the parabola is parallel to the negative y-axis.

Therefore, the maxima of the function is at (3,14). (Answer)

 The vertex form of the given function is y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{53}{4} and its axis of symmetry is x=-\dfrac{3}{2}.

Step-by-step explanation:  The given quadratic function is

y=5x^2+15x-2~~~~~~~~~~~~~~(i)

We are to rewrite the above function in vertex form and to determine its axis of symmetry.

We have from equation (i),

y=5x^2+15x-2\\\\\Rightarrow y=5(x^2+3x)-2\\\\\Rightarrow y=5\left(x^2+2\times x\times \dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{45}{4}-2\\\\\Rightarrow y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{45+8}{4}\\\\\Rightarrow y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{53}{4}.

So, the given function is a parabola with vertex at the point \left(-\dfrac{3}{2},-\dfrac{53}{4}\right).

Therefore, the axis of symmetry is given by

x=-\dfrac{3}{2}.

Thus, the vertex form of the given function is y=5\left(x+\dfrac{3}{2}\right)^2-\dfrac{53}{4} and its axis of symmetry is x=-\dfrac{3}{2}.

5x^2+15x-2
5(x^2+3x)-2
5(x^2+3x+2.25-2.25)-2
5(x^2+3x+2.25)-13.25
5(x+1.5)^2-13.25

final y=5(x+1.5)^2-13.25
axis of symmetry: (1.5, 13.25)


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