Question

How do you graph the parabola $y = {x^2} - 3x + 8$ using vertex, intercepts and additional points?

Answer 1

This problem deals with the conic sections. A conic section is a curve obtained as the intersection of the surface of a cone with a plane. There are three such types of conic sections which are, the parabola, the hyperbola and the ellipse. This problem is regarding one of those conic sections, which is a parabola. The general form of an equation of a parabola is given by ${x^2} = 4ay$.

Complete step-by-step answer:
The given equation is $y = {x^2} - 3x + 8$, the graph of the given equation can be obtained.
The equation of the curve looks like a parabola, a parabola has a vertex.
If the parabola is given by $y = a{x^2} + bx + c$, then the x-coordinate of the vertex is given by:
$\Rightarrow x = \dfrac{{ - b}}{{2a}}$
Here in the given parabola equation $y = {x^2} - 3x + 8$, here $a = 1,b = - 3$ and $c = 8$.
Now finding the x-coordinate of the vertex:
$\Rightarrow x = \dfrac{{ - \left( { - 3} \right)}}{{2\left( 1 \right)}}$
$\Rightarrow x = \dfrac{3}{2}$
Now to get the y-coordinate of the vertex of the parabola, substitute the value of $x = \dfrac{3}{2}$, in the parabola equation, as shown below:
$\Rightarrow y = {x^2} - 3x + 8$
$\Rightarrow y = {\left( {\dfrac{3}{2}} \right)^2} - 3\left( {\dfrac{3}{2}} \right) + 8$
Simplifying the above equation, as given below:
$\Rightarrow y = \dfrac{9}{4} - \dfrac{9}{2} + 8$
$\Rightarrow y = \dfrac{{23}}{4}$
So the vertex of the parabola $y = {x^2} - 3x + 8$ is A, which is given by:
$\Rightarrow A = \left( {\dfrac{3}{2},\dfrac{{23}}{4}} \right)$
Now the parabola $y = {x^2} - 3x + 8$ intersects the y-axis at $x = 0$, as given below:
$\Rightarrow y = {\left( 0 \right)^2} - 3\left( 0 \right) + 8$
$\therefore y = 8$
So the parabola $y = {x^2} - 3x + 8$ has an intercept at the point $\left( {0,8} \right)$.

Final Answer: The vertex of the parabola $y = - 3{x^2} + 5x - 2$ are $\left( {\dfrac{3}{2},\dfrac{{23}}{4}} \right)$ and $\left( {0,8} \right)$ respectively.

Note:
Please note that if the given parabola is${x^2} = 4ay$, then the vertex of this parabola is the origin $\left( {0,0} \right)$, and there is no intercept for this parabola as there are no terms of x or y. If the equation of the parabola includes any terms of linear x or y, then the vertex of the parabola is not the origin, the vertex has to be found by simplifying it into its particular standard form.