Chapter 9.1, Problem 77PFA

To determine

To evaluate the y -intercept of the quadratic equation.

Answer to Problem 77PFA

The y -intercept is at $(0,3)$

Explanation of Solution

Given information :

The quadratic equation is $f(x)=x^2-2x+3$

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

Highest point (if $a<0$ ) and be called the maximum. Here, the graph opens downwards.

Or, lowest point (if $a>0$ ) and can be called the minimum. Here the graph opens upwards.

In this case, ‘a’ is lesser than 0 hence the graph will have a maximum and will open downwards.

A parabola always points to infinity, either negative or positive. In a quadratic equation signifying parabola, the y -intercept is c.

Here, according to the given quadratic equation $c=3$ . Therefore, y -intercept is at $(0,3)$ .