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Rewriting quadratics in standard form

In [link] , the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.

Given a quadratic function, find the x - intercepts by rewriting in standard form .

  1. Substitute a and b into h = b 2 a .
  2. Substitute x = h into the general form of the quadratic function to find k .
  3. Rewrite the quadratic in standard form using h and k .
  4. Solve for when the output of the function will be zero to find the x - intercepts.

Finding the x -intercepts of a parabola

Find the x - intercepts of the quadratic function f ( x ) = 2 x 2 + 4 x 4.

We begin by solving for when the output will be zero.

0 = 2 x 2 + 4 x 4

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

f ( x ) = a ( x h ) 2 + k

We know that a = 2. Then we solve for h and k .

h = b 2 a k = f ( −1 ) = 4 2 ( 2 ) = 2 ( −1 ) 2 + 4 ( −1 ) 4 = −1 = −6

So now we can rewrite in standard form.

f ( x ) = 2 ( x + 1 ) 2 6

We can now solve for when the output will be zero.

0 = 2 ( x + 1 ) 2 6 6 = 2 ( x + 1 ) 2 3 = ( x + 1 ) 2 x + 1 = ± 3 x = 1 ± 3

The graph has x -intercepts at ( −1 3 , 0 ) and ( −1 + 3 , 0 ) .

We can check our work by graphing the given function on a graphing utility and observing the x - intercepts. See [link] .

Graph of a parabola which has the following x-intercepts (-2.732, 0) and (0.732, 0).
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In a Try It , we found the standard and general form for the function g ( x ) = 13 + x 2 6 x . Now find the y - and x -intercepts (if any).

y -intercept at (0, 13), No x - intercepts

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Applying the vertex and x -intercepts of a parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H ( t ) = 16 t 2 + 80 t + 40.

  1. When does the ball reach the maximum height?
  2. What is the maximum height of the ball?
  3. When does the ball hit the ground?
  1. The ball reaches the maximum height at the vertex of the parabola.
    h = 80 2 ( −16 ) = 80 32 = 5 2 = 2.5

    The ball reaches a maximum height after 2.5 seconds.

  2. To find the maximum height, find the y - coordinate of the vertex of the parabola.
    k = H ( b 2 a ) = H ( 2.5 ) = −16 ( 2.5 ) 2 + 80 ( 2.5 ) + 40 = 140

    The ball reaches a maximum height of 140 feet.

  3. To find when the ball hits the ground, we need to determine when the height is zero, H ( t ) = 0.

    We use the quadratic formula.

    t = −80 ± 80 2 4 ( −16 ) ( 40 ) 2 ( −16 ) = −80 ± 8960 −32

    Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

    t = 80 8960 32 5.458 or t = 80 + 8960 32 0.458

    The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See [link] .

    Graph of a negative parabola where x goes from -1 to 6.

    Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph.

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A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation H ( t ) = −16 t 2 + 96 t + 112.

  1. When does the rock reach the maximum height?
  2. What is the maximum height of the rock?
  3. When does the rock hit the ocean?

  1. 3 seconds
  2. 256 feet
  3. 7 seconds

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Questions & Answers

12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
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Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
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salma
Commplementary angles
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Uday
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salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
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Practice Key Terms 7

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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