For the function, f(x) = -5x² - 10x + 6,

find the vertex and y-intercept.

find the vertex and y-intercept.

The function, f(x) = -5x² - 10x + 6, is in standard form.

The y-intercept is the constant term (the number without a variable), because at the y-intercept, the value of x is equal to zero.

The x-coordinate of the vertex is

-b

x = -----

2a

In f(x) = -5x² - 10x + 6

a = -5, b = -10, c = 6

-(-10)

x = ------- {substituted -10 in for b and -5 in for a}

2(-5)

x = -1 {simplified}

The x-coordinate of the vertex is -1.

To find the y-coordinate of the vertex, substitute the x-coordinate, -1, back into the function.

f(x) = -5x² - 10x + 6

= -5(-1)² - 10(-1) + 6 {substituted -1 in for x}

= -5 + 10 + 6 {evaluated exponent and simplified}

= 11 {simplified}

The y-coordinate of the vertex is 11, which is also known as the maximum value of this particular function.

The y-intercept is the constant term (the number without a variable), because at the y-intercept, the value of x is equal to zero.

**The y-intercept is 6.**The x-coordinate of the vertex is

-b

x = -----

2a

In f(x) = -5x² - 10x + 6

a = -5, b = -10, c = 6

-(-10)

x = ------- {substituted -10 in for b and -5 in for a}

2(-5)

x = -1 {simplified}

The x-coordinate of the vertex is -1.

To find the y-coordinate of the vertex, substitute the x-coordinate, -1, back into the function.

f(x) = -5x² - 10x + 6

= -5(-1)² - 10(-1) + 6 {substituted -1 in for x}

= -5 + 10 + 6 {evaluated exponent and simplified}

= 11 {simplified}

The y-coordinate of the vertex is 11, which is also known as the maximum value of this particular function.

**The vertex is (-1,11)**

**- Algebra House**