y=x^2-14x+24
=x^2+16x+28
Identify where y is 0.
23. f(x)=\frac{1}{3}(x+5), f(x)=\frac{1}{9} x^2+\frac{0}{9} x-\frac{32}{9}
(25)a. The two equations f(x)=\frac{1}{3} x^2+\frac{10}{3} x-\frac{2}{3}
h=-16 t^2+700 are h=-16 t^2+300 and
b. 0=-16 t^2+700.
t \approx 6.6 ; 6.6-300, t \approx 4.3 ; 0=-16 t^2+700, 27a. The graph of g(x) \approx 2.3 seconds yards right and 20(x) is the graph of f(x) translated 200 reflected across the yards up, compressed vertically, and +20 29. Transle $x$-axis 27b. h(x)=0.0005(x-230)^2 the right 2 unilate the graph of f(x) up 7 units and to graph is comits. 31b. For f(x)=x^2, when b>1, the when 0<0 m p r e s s e d horizontally. 31c. For f(x)=x^2,
31d. For f(x), the graph is stretched horizontally. the neg f(x)=x^2, when b<0, the squared term offsets This egative. For example, when b=-3,(-3 x)^2=9 x^2. The is the same transformation as when b=3. Therefore, the sign of b has no effect on the transformation of this function.
33. Graph 1: y=-\frac{1}{3} x^2; Graph 2: y=-2 x^2. 35. The graph of y=3 x^2 is narrower than the graph of y=\frac{1}{2} x^2.
37. Both graphs have the same shape, but the graph of y=-3 x^2 opens down while the graph of y=3 x^2 opens up. 39. Both graphs have the same shape, but the graph of y=x^2+2 is translated up 2 units from the graph of y=x^2 and the graph of y=x^2-1 is translated down 1 unit from the graph of y=x^2.
Answer (1)
To find where the function y = x 2 − 14 x + 24 equals zero, we need to solve the equation
x 2 − 14 x + 24 = 0.
This is a quadratic equation, and we can solve it using the quadratic formula, factoring, or completing the square. In this case, let's try factoring.
Look for two numbers that multiply to 24 and add to -14. The numbers that satisfy this condition are -12 and -2 because:
− 12 × − 2 = 24
− 12 + ( − 2 ) = − 14
So, we can factor the quadratic as:
( x − 12 ) ( x − 2 ) = 0.
To find the values of x where y = 0 , set each factor equal to zero:
x − 12 = 0 which implies x = 12.
x − 2 = 0 which implies x = 2.
Therefore, the function y = x 2 − 14 x + 24 is zero at x = 12 and x = 2 .
For the second equation mentioned, y = x 2 + 16 x + 28 , you would follow a similar process to determine where y = 0 . However, let's focus on solving the first equation as given in the problem statement above. If you have a specific case you need help with from this second equation, feel free to ask!
Both of these are standard questions in High School algebra classes, typically under the topic of solving quadratic equations by factoring.
