Question

Quadratic below
2^2−7−4
Vertex -
Y-Intercept -
X-Intercept -

Answer 1

The y-intercept is y = -4.

The x-intercepts are
`x = 4` and
`x = -(1)/(2)`

The vertex is
`((7)/(4),-(81)/(8))`

Explanation:

Quadratic equation:

Has the following format:

`y = ax^2 + bx + c`

The y-intercept is c.

Finding the x-intercepts:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\Delta))/(2*a)`

`x_(2) = (-b - √(\Delta))/(2*a)`

`\Delta = b^(2) - 4ac`

Vertex:

Suppose we have a quadratic function in the following format:

`f(x) = ax^(2) + bx + c`

It's vertex is the point
`(x_(v), y_(v))`

In which

`x_(v) = -(b)/(2a)`

`y_(v) = -(\Delta)/(4a)`

Where

`\Delta = b^2-4ac`

In this question:

The quadratic equation is
`2x^2 - 7x - 4`, which has
`a = 2, b = -7, c = -4`. This means that the y-intercept is y = -4.

x-intercepts:

`\Delta = (-7)^2-4(2)(-4) = 81`

`x_(1) = (-(-7) + √(81))/(2*2) = 4`

`x_(2) = (-(-7) - √(81))/(2*2) = -(1)/(2)`

The x-intercepts are
`x = 4` and
`x = -(1)/(2)`

Vertex:

`x_(v) = -((-7))/(2*2) = (7)/(4)`

`y_(v) = -(81)/(4(2)) = -(81)/(8)`

The vertex is
`((7)/(4),-(81)/(8))`