What point is symmetrical to the y intercept of y=x^2-6x-40

1 answer

To find the point that is symmetrical to the y-intercept of the function \( y = x^2 - 6x - 40 \), we first need to determine the y-intercept itself.

The y-intercept occurs when \( x = 0 \). Plugging \( x = 0 \) into the equation:

\[ y = 0^2 - 6(0) - 40 = -40 \]

So, the y-intercept is the point \( (0, -40) \).

Next, we need to find the axis of symmetry of the parabola represented by the equation. The axis of symmetry for a quadratic equation of the form \( y = ax^2 + bx + c \) can be found using the formula:

\[ x = -\frac{b}{2a} \]

In our case, \( a = 1 \) and \( b = -6 \):

\[ x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3 \]

The axis of symmetry is the vertical line \( x = 3 \).

To find the point symmetrical to the y-intercept \( (0, -40) \), we mirror this point across the axis of symmetry \( x = 3 \). To do that, we need to find the horizontal distance from the y-intercept to the axis of symmetry:

\[ \text{Distance} = 3 - 0 = 3 \]

Now, to find the symmetrical point, we move the same distance to the right of the axis of symmetry:

\[ 3 + 3 = 6 \]

The y-coordinate remains the same since we’re reflecting horizontally. Thus, the symmetrical point is given by:

\[ (6, -40) \]

Therefore, the point symmetrical to the y-intercept of the quadratic function is \( (6, -40) \).