Interpreting Quadratic Functions in Standard Form

To draw the graph of the given quadratic function written in standard form, we must start by identifying the values of $a,$ $b,$ and $c.$ $$\begin{array}{c}y=2x^2-6x+1\iff y=2x^2+(\text{-}6x)+1\end{array}$$ We can see that $a=2,$ $b=\text{-}6,$ and $c=1.$ Now, we will follow four steps to graph the function.

1. Find the axis of symmetry.
2. Calculate the vertex.
3. Identify the $y\text{-}$intercept and its reflection across the axis of symmetry.
4. Connect the points with a parabola.

Finding the Axis of Symmetry

The axis of symmetry is a vertical line with equation $x=\text{-}\frac{b}{2a}.$ Since we already know the values of $a$ and $b,$ we can substitute them into the formula. 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 $x\text{-}$ and $y\text{-}$coordinates in terms of $a$ and $b.$ $$\begin{array}{c}\text{Vertex:}\left(\text{-}\frac{b}{2a},f\left(\text{-}\frac{b}{2a}\right)\right)\end{array}$$ Note that the formula for the $x\text{-}$coordinate is the same as the formula for the axis of symmetry, which is $x=1.5.$ Thus, the $x\text{-}$coordinate of the vertex is also $1.5.$ To find the $y\text{-}$coordinate, we need to substitute $1.5$ for $x$ in the given equation. We found the $y\text{-}$coordinate, and now we know that the vertex is $(1.5,\text{-}3.5).$

Identifying the $y\text{-}$intercept and its Reflection

The $y\text{-}$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 $y\text{-}$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.