Question #010fe

1 Answer
Jun 26, 2017

Transform the equation to factored form, and answer the following questions.

Explanation:

Basically, what they're asking is to transform the equation to vertex form, and answer the following questions using the now-transformed equation.

f(x)=1/2x^2+1/2x-10


Intercept Form

Intercept form is another way to say factored form. This involves factoring.

First, let's factor out the GCF - 1/2.

f(x)=1/2x^2+1/2x-10

f(x)=1/2(x^2+x-20)

Now it's a simple trinomial. "What two numbers added equal b, and multiplied equal c?"

In this case, it is 5 and -4.

f(x)=1/2(x+5)(x-4)

This is the factored form (AKA intercept form).

If we expand the brackets, we will get the standard form.


Zeros

In order to find the x-intercepts, the most simplest method is to use the quadratic formula.

x = (-b+-sqrt(b^2-4ac))/(2a)

Here, we sub in the values from a standard equation.

x = (-(1/2)+-sqrt((1/2)^2-4(1/2)(-10)))/(2(1/2))

We simplify.

x = (-1/2+-sqrt(81/4))/(1)

And solve for x.

x [+] = 4

x [-] = -5

If we graph it, we can conclude that the zeros are 4 and -5.


Vertex

To find the vertex, in my opinion, the easiest way is to complete the square.

First, we factor out the a-value from the first two terms.

f(x)=1/2x^2+1/2x-10

f(x)=1/2(x^2+x)-10

Now, we divide the b-value by two, and the square it. In the end, we still get one. We must add and subtract those values in order to keep the equation true to itself.

f(x)=1/2(x^2+x + 1/4 - 1/4)-10

Now we expand the -1/4.

f(x)=1/2(x^2+x + 1/4) -1/8 -10

Now we simplify the equation.

f(x)=1/2(x + 1/2)^2 -81/8

That's our vertex form. If we expand it, we would get the standard form.

Since we have our vertex formula, we can just use the h and the k-values where (h,k) = (x, y).

Thus, the vertex is (-1/2, -81/2).

If we graph it, we can conclude the that vertex is indeed (-1/2, -81/2).


Axis of Symmetry

The axis of symmetry is basically the x-value of the vertex.

Thus, it is: x=-1/2.


Here's a graph.

graph{1/2x^2+1/2x-10 [-10, 10, -5, 5]}


Hope this helps :)