standard form of -16(t+1)(t-5)

Answer 1

When a question asks for standard form they want you to order the variables from the highest degree (biggest exponent) to the lowest degree followed by any constant term (a number with no exponent). Since your problem only has one variable you don't need to worry about degrees just yet; you just need to multiply out the expression and order them by highest exponent to the lowest.

-16(t+1)(t-5) → use the Foil method

First: t × t = t²

Outer: t × 5 = 5t

Inner: 1 × t = 1t

Last: 1 x -5 = - 5

This gives us: t² + 5t + 1t - 5 → t² + 6t - 5

So we plug that back in and multiply by -16

-16( t² + 6t - 5)

-16t² - 96t + 256

This is already in standard form! we don't have to change the order!

Answer 2

please see video

Answer 3

What is the standard form of a polynomial?

A polynomial is expressed in its standard form if the terms of the polynomial are written with their degree in decreasing order.

5x² + 2x is in standard form.

2x + 5x² is not in standard form, because the degree is increasing from 1 to 2 as we go along.

If we take the equation given in the question, we can put it in the standard form doing the multiplication:

-16(t+1)(t-5) = -16[t(t-5)+(t-5) ]= -16(t²-5t+1-5) = -16(t²-4t-5) = -16t²+64t+80

Pay attention to your signs when you multiply. It is easy to forget a minus sign and end up with an incorrect value. Also, in this case, my answer was already in the standard form. However, there will be times when your answer may not be in the standard form, and you'd then have to reorder your terms to put them in the standard form.

Answer 4

I have solved this question in this video for you. Please check it out and let me know if you have any additional questions.

Take care,

Dr. Christal-Joy Turner

Answer 5

= -16(t^2 -4t -5)

= -16t^2 +64t+ 80

= standard form with ax^2 +bx+c where a=-16,b=64, c=80

complete the square

= -16(t^2-4t+4)+80+64

= -16(x-2)^2 +144

= vertex form with vertex = maximum point= (2,144)

=a(x-h)^2 + k where (h,k) = vertex = (2,144), h=2, k=144, a<0 indicating a downward opening parabola

take the derivative, set =0, solve for x

y'=-32x+64=0

x=64/2=2

x=2 is the axis of symmetry of the parabola

y(2)=144 = maximum

x intercepts =(-1,0),(5,0)= -1, 5= zeros, roots, or solutions

y intercept =(0,80)= 80

(4,80) is another point on the parabola, reflected over the axis of symmetry

that gives you 5 points to sketch the graph