Question

How do you graph the parabola $y = {x^2} - 4x + 7$ using vertex, intercepts, and additional points?

Answer 1

Hint: In this problem, a quadratic equation is given. The equation of the curve is a parabola, and it has a vertex. In the general form, the equation is $y = a{x^2} + bx + c$, then the x-coordinate of the vertex is given by $x = - \dfrac{b}{{2a}}$. Then substitute the value of x in the given quadratic equation to find the y-coordinate. To find y-intercept by putting x is equal to 0 in the given quadratic equation. Therefore, we will find the two points of the parabola.

Complete step-by-step solution:
In this question, the quadratic equation is given. The graph of the given expression is obtained.
Here, in the given parabola equation is:
$ \Rightarrow y = {x^2} - 4x + 7$ ……………....(1)
Let us compare the above equation with the general form $y = a{x^2} + bx + c$.
So, the value of ‘a’ is 1, the value of b is ‘-4’, and the value of ‘c’ is 7.
Now, let us find the x-coordinate of the vertex.
$ \Rightarrow x = - \dfrac{b}{{2a}}$
Substitute the values in the above equation.
$ \Rightarrow x = - \dfrac{{\left( { - 4} \right)}}{{2\left( 1 \right)}}$
Let us simplify the equation.
$ \Rightarrow x = \dfrac{4}{2}$
That is equal to,
$ \Rightarrow x = 2$
Now, substitute the value of x in the equation (1).
$ \Rightarrow y = {x^2} - 4x + 7$
Put x is equal to 2.
$ \Rightarrow y = {\left( 2 \right)^2} - 4\left( 2 \right) + 7$
Let us simplify the above equation.
$ \Rightarrow y = 4 - 8 + 7$
That is equal to,
$ \Rightarrow y = 3$
Therefore, the vertex of the parabola is (2,3).
Now, let us find the y-intercept by putting x is equal to 0 in the equation (1).
 $ \Rightarrow y = {\left( 0 \right)^2} - 4\left( 0 \right) + 7$
Let us simplify the right-hand side.
$ \Rightarrow y = 0 - 0 + 7$
That is equal to,
$ \Rightarrow y = 7$
So, the parabola has an intercept at the point (0,7).

Hence, the vertex of the parabola $y = {x^2} - 4x + 7$ are $(2,3)$ and $(0,7)$.

Note: Since the parabola equation includes linear x and y terms, then the vertex of the parabola can never be the origin. If the given parabola is in the form of ${x^2} = 4ay$, then the vertex of this parabola is the origin (0,0), and there is no intercept for this parabola as there are no terms of x or y.