Solveeit Logo
Login

Question

Question: How do you graph the parabola \(y = {x^2} - 4x + 7\) using vertex, intercepts, and additional points...

How do you graph the parabola y=x2−4x+7y = {x^2} - 4x + 7 using vertex, intercepts, and additional points?

Explanation

Solution

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=ax2+bx+cy = a{x^2} + bx + c, then the x-coordinate of the vertex is given by x=−b2ax = - \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:
⇒y=x2−4x+7\Rightarrow y = {x^2} - 4x + 7 ……………....(1)
Let us compare the above equation with the general form y=ax2+bx+cy = 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.
⇒x=−b2a\Rightarrow x = - \dfrac{b}{{2a}}
Substitute the values in the above equation.
⇒x=−(−4)2(1)\Rightarrow x = - \dfrac{{\left( { - 4} \right)}}{{2\left( 1 \right)}}
Let us simplify the equation.
⇒x=42\Rightarrow x = \dfrac{4}{2}
That is equal to,
⇒x=2\Rightarrow x = 2
Now, substitute the value of x in the equation (1).
⇒y=x2−4x+7\Rightarrow y = {x^2} - 4x + 7
Put x is equal to 2.
⇒y=(2)2−4(2)+7\Rightarrow y = {\left( 2 \right)^2} - 4\left( 2 \right) + 7
Let us simplify the above equation.
⇒y=4−8+7\Rightarrow y = 4 - 8 + 7
That is equal to,
⇒y=3\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).
⇒y=(0)2−4(0)+7\Rightarrow y = {\left( 0 \right)^2} - 4\left( 0 \right) + 7
Let us simplify the right-hand side.
⇒y=0−0+7\Rightarrow y = 0 - 0 + 7
That is equal to,
⇒y=7\Rightarrow y = 7
So, the parabola has an intercept at the point (0,7).

Hence, the vertex of the parabola y=x2−4x+7y = {x^2} - 4x + 7 are (2,3)(2,3) and (0,7)(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 x2=4ay{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.