(x-7)^2=0 ⇔ x-7=0
We have found the equivalent equation x-7=0. We can solve it with one step.
x-7=0 ⇔ x=7
Since the equation x-7=0 holds true for x=7, this is the solution for the original quadratic equation as well.
Now that we have our function in standard form, we can compare it to the general form to identify the parameters a and b. This will let us find the vertex by using the formula for the x-coordinate of the vertex. Let's identify the parameters needed.
ax^2+ bx+c
1x^2+( - 14)x+49
As we can see, a= 1 and b= -14. Now, we are ready to use the formula for the x-coordinate of the vertex.
V_x = - b/2a
Let's give it a try.
Hence, the solution for the quadratic equation (x+3)^2=0 is x= - 3. Furthermore, since the related function is y=(x+3)^2, we can rewrite it in standard by expanding the square binomial, just as we did in Part B. The result is shown below.
y=(x+3)^2 ⇔ y=x^2+6x+9
Now, comparing the function to the standard form we can identify the parameters a and b.
y = ax^2+ bx+c
y= 1x^2+ 6x+9
As we can see, a = 1 and b= 6. With this we are ready to find the x-coordinate of the vertex.
d Let's take a look at the results from Parts B and C.
Quadratic Function
Vertex
y = (x-7)^2
(7,0)
y = x+3
⇕
y = x-(- 3)
(- 3,0)
The results suggest that a quadratic function of the form y=(x-h)^2 has its vertex located at the point (h,0). Let's have a look at the function given.
y = (x +4)^2 ⇔ y = (x - (-4))^2
Notice that it is of the form y=(x-h)^2 as well.
y=(x- h)^2
y=(x-( -4))^2
Therefore, we can expect it to have its vertex at ( -4,0).
