To draw the graph of the given quadratic function written in standard form, we must start by identifying the values of a, b, and c.
y=2x^2-6x+1 ⇔ y=2x^2+(-6x)+1
We can see that a=2, b=-6, and c=1. Now, we will follow four steps to graph the function.
The axis of symmetry of the parabola is the vertical line with equation x=1.5.
Calculating the Vertex
To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b.
Vertex: ( - b/2a, f( - b/2a ) )
Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=1.5. Thus, the x-coordinate of the vertex is also 1.5. To find the y-coordinate, we need to substitute 1.5 for x in the given equation.
We found the y-coordinate, and now we know that the vertex is (1.5,-3.5).
Identifying the y-intercept and its Reflection
The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,1). Let's plot this point and its reflection across the axis of symmetry.
Connecting the Points
We can now draw the graph of the function. Since a=2, which is positive, the parabola will open upward . Let's connect the three points with a smooth curve.
