How do you find the minimum and maximum value of #y=(x+7)(x+3)#?

Question

Eli Assouline · Accepted Answer

minimum #->(x,y)=(-5,-4)#
maximum #-> y=+oo# If you multiply out the brackets you have the general form of: #y=ax^2+bx+c# in this case #a=1# giving just #+x^2# As this is positive the graph is of general shape #uu# Thus the vertex is a minimum. The vertex is #1/2# way between the x-intercepts. Set:#" "y=0=(x+7)(x+3)# Thus #x+7 =0 => x=-7# #x+3=0=>x=-3# So #x_("vertex")=(-3-7)/2=-5# Substitute #x=-5# to give: #y_("vertex")=(-5+7)(-5+3)= 2xx(-2)=-4# Thus the minimum #->(x,y)=(-5,-4)# As the graph is of form #uu# then the maximum is #y=+oo#

Eli Assouline · Answer

undefined To find the minimum and maximum values of $ y = (x + 7)(x + 3) $, you can use the properties of quadratic functions. First, expand the expression to get $ y = x^2 + 10x + 21 $. Since this is a quadratic function, its graph is a parabola. The minimum or maximum value occurs at the vertex of the parabola, which is given by the formula $ x = -\frac{b}{2a} $, where $ a $ and $ b $ are the coefficients of the quadratic equation $ ax^2 + bx + c $.

In this case, $ a = 1 $ and $ b = 10 $. Substituting these values into the formula, you get $ x = -\frac{10}{2(1)} = -5 $. Now, substitute $ x = -5 $ into the original equation to find the corresponding $ y $ value. $ y = (-5)^2 + 10(-5) + 21 = 25 - 50 + 21 = -4 $.

Therefore, the minimum value of $ y $ is $ -4 $, which occurs when $ x = -5 $. Since the coefficient of $ x^2 $ is positive, the parabola opens upwards, meaning that $ y = -4 $ is the minimum value.

To find the maximum value, you can consider the behavior of the parabola. Since the coefficient of $ x^2 $ is positive, the parabola opens upwards, and it extends infinitely upward. Hence, there is no maximum value for $ y $.