How do you find the vertex and the intercepts for #y = -x^2 - 4x + 7#?

Question

Eli Assouline · Accepted Answer

Vertex is at # (-2,11)#, y intercept is at #(0,7)#, x-interceps are at #(-5.32 , 0) and (1.32,0)# #y= -x^2-4x+7 or y = -(x^2+4x)+7 or y = -(x^2+4x +4)+4+7# or #y = -(x+2)^2+11 # comparing with standard equation #y=a(x-h)^2+k, (h,k)# being vertex we find here vertex at #h= -2 ,k=11) or (-2,11)# y-intercept can be found by putting #x=0# in the equation, #y= 7# y intercept is at #(0,7)# x-intercept can be found by putting #y=0# in the equation , #-(x+2)^2 +11 =0 or (x+2)^2 =11 or (x+2) = +-sqrt 11 :. x= -2+sqrt11 ~~ 1.32 (2dp) # or # x= -2 - sqrt11 ~~ -5.32 (2dp)# x-intercepts are at #(-5.32 , 0) and (1.32,0)# [Ans]

Isaiah Anthony · Answer

undefined To find the vertex of the quadratic function $ y = -x^2 - 4x + 7 $, you can use the formula $ x = \frac{{-b}}{{2a}} $, where $ a $ is the coefficient of the quadratic term, $ b $ is the coefficient of the linear term, and $ c $ is the constant term.

Using the formula, the x-coordinate of the vertex is:
$$ x = \frac{{-(-4)}}{{2 \cdot (-1)}} = \frac{4}{-2} = -2 $$

To find the y-coordinate of the vertex, substitute the x-coordinate into the equation:
$$ y = -( -2)^2 - 4(-2) + 7 = -4 - (-8) + 7 = 11 $$

So, the vertex of the parabola is $ (-2, 11) $.

To find the x-intercepts, set $ y = 0 $ and solve for $ x $:
$$ -x^2 - 4x + 7 = 0 $$
$$ x^2 + 4x - 7 = 0 $$

Using the quadratic formula:
$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Substituting the coefficients $ a = -1 $, $ b = 4 $, and $ c = -7 $:
$$ x = \frac{{-4 \pm \sqrt{{4^2 - 4(-1)(-7)}}}}{{2 \cdot (-1)}} $$
$$ x = \frac{{-4 \pm \sqrt{{16 - 28}}}}{{-2}} $$
$$ x = \frac{{-4 \pm \sqrt{{-12}}}}{{-2}} $$
$$ x = \frac{{-4 \pm 2i\sqrt{3}}}{{-2}} $$

Thus, the x-intercepts are complex and are not real numbers.

To find the y-intercept, set $ x = 0 $ and solve for $ y $:
$$ y = -(0)^2 - 4(0) + 7 = 7 $$

So, the y-intercept is $ (0, 7) $.