The constant of a function tells us the y-coordinate of its y-intercept.
y=2x^2+4x+ 2 ← constantThe y-intercept is (0,2). To find the x-intercepts, we set y equal to 0 and solve for x by using the Quadratic Formula
The equation has one x-intercept at (-1,0). Since there is only one x-intercept, this point must also be the parabola's locator point, or vertex.
Finally, we want to write the equation in graphing form.
Graphing Form:& y=a(x- h)^2+ k
Vertex:& ( h, k)
Let's substitute the function's vertex in this equation. Notice that the value of a equals the coefficient of x^2. From the equation we see that this is a= 2. With this information, we can write the complete equation.
Function:& y= 2(x-( -1))^2+ 0
Vertex:& ( -1, 0)
This simplifies to y=2(x+1)^2.
b Like in Part A, we will start by simplifying the equation and solving it for y.
The function has two x-intercepts, x=0 and x= 2. All parabolas are symmetric about their vertex. What this means is if two points have the same y-coordinate, such as the x-intercepts, they are equidistant from the parabola's line of symmetry. Therefore, we can find the line of symmetry by averaging the x-intercepts.
From the diagram, we see that the function has a vertex in (1,1). Also, the squared variable in the simplified equation has a coefficient of a= -1. With this information, we can write the function in graphing form.
Function:& y= -1(x- 1)^2+ 1
Vertex:& ( 1, 1)
This simplifies to y=-(x-1)^2+2
