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### Section 4-2 : Parabolas

6. Sketch the graph of the following parabola. The graph should contain the vertex, the $$y$$‑intercept, $$x$$-intercepts (if any) and at least one point on either side of the vertex.

$f\left( x \right) = 4{x^2} - 4x + 1$

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Start Solution

Let’s find the vertex first. In this case the equation is in the form $$f\left( x \right) = a{x^2} + bx + c$$. And so we know the vertex is the point $$\left( { - \frac{b}{{2a}},f\left( { - \frac{b}{{2a}}} \right)} \right)$$. The vertex is then,

$\left( { - \frac{{ - 4}}{{2\left( 4 \right)}},f\left( { - \frac{{ - 4}}{{2\left( 4 \right)}}} \right)} \right) = \left( {\frac{1}{2},f\left( {\frac{1}{2}} \right)} \right) = \left( {\frac{1}{2},0} \right)$

Also note that $$a = 4 > 0$$ for this parabola and so the parabola will open upwards.

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The $$y$$-intercept is just the point $$\left( {0,f\left( 0 \right)} \right)$$. A quick function evaluation gives us that $$f\left( 0 \right) = 1$$ and so for our equation the $$y$$-intercept is $$\left( {0,1} \right)$$.

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For the $$x$$-intercepts we would normally solve the equation $$f\left( x \right) = 0$$. However, in this case we don’t need to do that. From the first step we see that the vertex has a y‑coordinate of zero and hence is also an $$x$$-intercept. Also, because it is the vertex this can be the only $$x$$-intercept for this function.

Note that if we’d solved the equation we would have also arrived at this single $$x$$‑intercept.

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In this case all we have are the vertex (which also happens to be the single $$x$$‑intercept) and the $$y$$-intercept (which is on the right side of the vertex). So, we’ll need a point that is on the left side of the vertex and we can find the point on the left side of the vertex that corresponds to the $$y$$-intercept for this point.

The $$y$$-intercept is a distance of 1 to the left of the vertex and so there will be a corresponding point at the same $$y$$ value to the right and it will be a distance of 1 to the right of the vertex. Therefore, the point to the right of the vertex corresponding to the y‑intercept is $$\left( {1,1} \right)$$.

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Here is a sketch of the parabola including all the points we found above. 