how to identify the vertex and the x and y intercept of h(x)=4x^2-4x+21

Question

unklerhaukus · Answer

the y intercept is when x=0

the x intercept is when y=0

the vertex is when the first derivative is zero

anonymous · Answer

dont you have to change it into standard form, h(x)=(x-k)+h, first?

unklerhaukus · Answer

dont do that

anonymous · Answer

why?

unklerhaukus · Answer

the equation is already in standard form

unklerhaukus · Answer

$$h(x)=4x^2-4x+21$$
lets find the  $y$-intercept

setting $x=0$
$$h(0)=4(0)^2-4(0)+21$$
$$h(0)=$$

anonymous · Answer

so it's 21

anonymous · Answer

but how to find the vertex??

unklerhaukus · Answer

the $y$-intercept is $(0,21)$

unklerhaukus · Answer

$$h(x)=4x^2-4x+21$$
$$h'(x)=\qquad\qquad\qquad=0$$

anonymous · Answer

ummm so would i do complete the square?

unklerhaukus · Answer

take the first derivative

anonymous · Answer

haha im sorry but what do you mean when you say derivative??

unklerhaukus · Answer

oh, ok. well have to find the vertex a different way then

unklerhaukus · Answer

for
$$h(x)=ax^2+bx+c$$
the x-coordinate is at $$x=\frac{-b}{2a}$$

anonymous · Answer

i got 1/2

unklerhaukus · Answer

what is the y coordinate of that point?

anonymous · Answer

ummm im not sure how to find it...

unklerhaukus · Answer

you found the x coordinate of the vertex, is 1/2
 to find the y coordinate
substitute this into the equation

h(x)=4x^2-4x+21

h(1/2)=4(1/2)^2+4(1/2)+21

anonymous · Answer

i got 20

anonymous · Answer

Thank you very much for the help.