Explanation:
if you don't understand 14., then how did you solve 13. ?
14.
first, remember a few things about a quadratic function (or parabola) :
y = (fx) = ax² + bx + c
the function curve is statically going with increasing x from ±infinity to an extreme point (vertex), where the curving tendency (slope) is turning around, and then the curve continues back to ±infinity.
if the value of “a” is positive, the parabola opens upward (and the vertex is a minimum as the curve goes from +infinity to +infinity), if the value is negative it opens downward (and the vertex is a maximum as the curve goes from -infinity to -infinity).
the x-coordinate (h) of the vertex is
h = -b/2a
the y-coordinate (k) of the vertex is simply
k = f(h)
FYI : there is also the possibility of
x = ay² + by + c
for the parabola to open left or right, but that is not the case here.
let's have a look at our parabola here :
y = f(x) = x² + x - 4
a = 1 (positive, so, it is opening up)
b = 1
c = -4
h = -1/(2×1) = -1/2 = -0.5
k = (-0.5)² + (-0.5) - 4 = 0.25 - 0.5 - 4 = -4.25
so, to answer the question, we know that with growing x right of x = -0.5, f(x) is continually increasing. there is no strange up/down wobbling. if one x-value (> -0.5) is larger than another x-value, then also the corresponding y-value is larger than the other y-value.
in the same way we know that with growing x left of x = -0.5, f(x) is continually decreasing. there is again no strange up/down wobbling. if one x-value (< -0.5) is larger than another x-value, then the corresponding y-value is smaller than the other y-value.
so, what is the smallest valid y-value ?
y = f(x) = -2
for what x-value do we get f(x) = -2 ?
-2 = x² + x - 4
0 = x² + x - 2
it is simple, and with some experience by just looking at it we know
x1 = 1
x2 = -2
but formally we need to solve a quadratic equation
0 = ax² + bx + c
in our case here
a = 1
b = 1
c = -2
remember the general solution
x = (-b ± sqrt(b² - 4ac))/(2a) =
= (-1 ± sqrt(1² - 4×1×-2))/(2×1) =
= (-1 ± sqrt(1 + 8))/2 = (-1 ± sqrt(9))/2 =
= (-1 ± 3)/2
x1 = (-1 + 3)/2 = 2/2 = 1
x2 = (-1 - 3)/2 = -4/2 = -2
and now, what is the largest valid y-value ?
y = f(x) = 16
for what x-value do we get f(x) = 16 ?
16 = x² + x - 4
0 = x² + x - 20
again, we need to solve a quadratic equation.
in our case here
a = 1
b = 1
c = -20
x = (-b ± sqrt(b² - 4ac))/(2a) =
= (-1 ± sqrt(1² - 4×1×-20))/(2×1) =
= (-1 ± sqrt(1 + 80))/2 = (-1 ± sqrt(81))/2 =
= (-1 ± 9)/2
x1 = (-1 + 9)/2 = 8/2 = 4
x2 = (-1 - 9)/2 = -10/2 = -5
so, we have the possibilities of the points
(1, -2) and (-2, -2)
(4, 16) and (-5, 16)
is now the the combination of
(1, -2) and (4, 16) satisfying the condition
a <= x <= a+3
yes, we are on the right side of the vertex, for a = 1 we get 1 <= x <= 4
for increasing x between 1 and 4 we get increasing f(x)-values between -2 and 16.
is
(-2, -2) and (-5, 16) satisfying the condition ?
as we are left of the vertex, we actually need to look at them this way :
(-5, 16) and (-2, -2)
yes, we are on the left side of the vertex, for a = -5 we get -5 <= x <= -2.
for increasing x between -5 and -2 we get decreasing f(x)-values between 16 and -2.
between x=1 and x=-5 or x=-2 and x=4 or x=-5 and x=4 or x=4 and x=-5 the difference is higher than 3, so they are not valid combinations.
between x=-2 and x=1 we don't reach f(x)=16 anywhere.
between x=1 and x=-2 or x=-2 and x=-5 the direction is wrong to satisfy the condition.
so, the only possible solutions are
a = 1 and a = -5