# How to find the x and y-intercepts of a quadratic function.

**Solution:**

We will use the concept of a quadratic equation to find the x and y-intercept of the quadratic equation.

Let us see how we will use the concept of a quadratic equation to find the x and y-intercept of the quadratic equation.

Let us consider the general form of quadratic equation as y = ax^{2 }+ bx + c

In order to find the x-intercept, we will make the y coordinates of the curve 0 because the x-intercept is the point where the curve cuts the x-axis and at the x-axis, the coordinate of y is equal to 0.

Hence, keeping y = 0 , we get ax^{2 }+ bx + c = 0

We have to solve the quadratic equation to find the intercepts on the x-axis.

However, using Sridharacharya's formula we get;

x = [ - b + √(b^{2} - 4ac) ] / 2a, [ - b - √(b^{2} - 4ac) ] / 2a which are the intercepts on the x-axis.

Similarly in order to find the y-intercept we have to make x coordinates = 0 because y-intercept is the point where the graph cuts the y-axis and hence the x coordinate on the y-axis is equal to 0.

Hence, putting x = 0 in the general form of the quadratic equation we get;

y = c is the intercept on the y-axis.

Thus, using Sridharacharya's formula we get x = [ - b + √(b^{2} - 4ac) ] / 2a, [ - b - √(b^{2} - 4ac) ] / 2a which are the intercepts on the x-axis and y = c is the intercept on the y-axis.

## How to find the x and y-intercepts of a quadratic function.

**Summary:**

y = c, is the intercept on y-axis, using Sridharacharya's formula we get x = [ - b + √(b^{2}- 4ac) ] / 2a, [ - b - √(b- 4ac) ] / 2a are the intercepts on the x axis.

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