Find the vertex form for the following quadratic function. f(x)=x^2+14x+41

f(x) = (x + 7)² - 8

the equation of a quadratic in vertex form is

y = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier

given the quadratic in standard form : y = ax² + bx + c ( a ≠ 0 )

then the x-coordinate of the vertex is

= -

f(x) = x² + 14x + 41 is in standard form

with a = 1, b = 14 and c = 41

= - = - 7

substitute this value into the equation for y- coordinate

y = (- 7 )² + 14(- 7 ) + 41 = 49 - 98 + 41 = - 8

f(x) = (x + 7)² - 8 ← in vertex form

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc