I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 1.10 : Common Graphs
11. Without using a graphing calculator sketch the graph of \(h\left( x \right) = 2{x^2} - 3x + 4\).
We know that the graph of \(f\left( x \right) = a{x^2} + bx + c\) will be a parabola that opens upwards if \(a > 0\) and opens downwards if \(a < 0\). We also know that its vertex is at,
\[\left( { - \frac{b}{{2a}},f\left( { - \frac{b}{{2a}}} \right)} \right)\]The \(y\)-intercept of the parabola is the point \(\left( {0,f\left( 0 \right)} \right) = \left( {0,c} \right)\) and the \(x\)-intercepts (if any) are found by solving \(f\left( x \right) = 0\).
So, or our case we know we have a parabola that opens upwards and that its vertex is at,
\[\left( { - \frac{{ - 3}}{{2\left( 2 \right)}},f\left( { - \frac{{ - 3}}{{2\left( 2 \right)}}} \right)} \right) = \left( {\frac{3}{4},f\left( {\frac{3}{4}} \right)} \right) = \left( {\frac{3}{4},\frac{{23}}{8}} \right) = \left( {0.75,\,\,2.875} \right)\]We can also see that the \(y\)-intercept is \(\left( {0,4} \right)\). Because the vertex is above the \(x\)-axis and the parabola opens upwards we can see that there will be no \(x\)-intercepts.
It is usually best to have at least one point on either side of the vertex and we know that parabolas are symmetric about the vertical line running through the vertex. Therefore, because we know that the \(y\)-intercept is 0.75 units to the left of the vertex that we must also have a point that is 0.75 to the right of the vertex with the same \(y\)-value and this point is : \(\left( {1.5,4} \right)\).
Here is a sketch of this parabola.
