684 Worked-Out Solutions to Exercises: Chapters 21 to 29
- Here’s the quadratic function again, for reference:
y=− 2 x^2 + 2 x− 5The coefficient of x^2 is negative. Therefore, when we graph the function, we get a parab-
ola that opens downward.- To find the real zeros (if any), we can calculate the discriminant based on the general
standard polynomial equation for a quadratic:
y=ax^2 +bx+cIn this situation, we have a=−2,b= 2, and c=−5. The discriminant, d, isd=b^2 − 4 ac
= 22 − 4 × (−2)× (−5)
= 4 − 40
=− 36Because d is negative, we know that this function has no real zeros.- The x-coordinate of the vertex point, which represents an absolute maximum because
the parabola opens downward, is
xmax=−b/(2a)
=−2 / [2 × (−2)]
=−2 / (−4)
= 2/4
= 1/2They-coordinate can be found by plugging in xmax and calculating from the function:ymax=− 2 x^2 max+ 2 xmax− 5
=− 2 × (1/2)^2 + 2 × 1/2 − 5
=− 2 × 1/4 + 1 − 5
=−1/2+ 1 − 5
= 1/2 − 5
=−9/2The coordinates of the vertex are (1/2, −9/2).