How do you write #h= -0.04(d-19)^2 +14.44# in factored form?

Question

Isaiah Anthony · Accepted Answer

The factored form is #h=-0.04d(d-38)#

The equation as given in the problem statement is in vertex form. Factored form requires knowledge of the roots of the equation. It is easy to get the roots from the vertex form. Simply set the vertex form = 0.

#-0.04(d-19)^2+14.44=0#

Subtract 14.44 from both sides and divide both sides by -0.04.

#(d-19)^2=(-14.44)/(-0.04)=361=19^2#

Take the square root of both sides.

#d-19=+-19#

Add 19 to both sides.

#d=19+-19#

So the two roots are

#r_1=0#, and

#r_2=38#.

The factored form of a quadratic equation is

#h=a(d-r_1)(d-r_2)#

Here, a = -0.04, so in factored form we would have

#h=-0.04(d-0)(d-38)=-0.04d(d-38)#

We can check this algebraically by converting the vertex form and the factored form into the standard form and see if we get the same thing.

CONVERT FROM VERTEX TO STANDARD FORM

#h=-0.04(d-19)^2+14.44=-0.04(d^2-38d+361)+14.44#

#h=-0.04d^2+1.52d-14.44+14.44=-0.04d^2+1.52d+0#

CONVERT FROM FACTORED to STANDARD FORM

#h=-0.04d(d-38)=-0.04d^2+1.52d+0#

The standard form is the same, so we have the correct answer.

Isaiah Anthony · Answer

undefined To write $ h = -0.04(d-19)^2 +14.44 $ in factored form, you would first factor out the common factor from the quadratic term, then rewrite it in squared form.

$ h = -0.04(d-19)^2 +14.44 $  
$ h = -0.04 \times (d-19)^2 +14.44 $  
$ h = -0.04 \times (d-19)(d-19) +14.44 $  
$ h = -0.04 \times (d-19)(d-19) +14.44 $  
$ h = -0.04(d-19)(d-19) +14.44 $