SAT · Math · Advanced Math
Funciones cuadráticas y exponenciales: evaluarlas, encontrar máximos o mínimos, traslaciones e interpretar tasas de crecimiento.
$f(x)=-(x-4)2+9$: el vértice es $(4, 9)$, así que el valor máximo es 9 (y se alcanza en $x=4$).
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
The graph of $y=f(x)$ is shown. What is the minimum value of $f(x)$?
Respuesta numérica (escribes el número).
Respuesta: -3
The minimum value of $f(x)$ is the $y$-coordinate of the lowest point of the graph. The lowest point (the vertex) is $(2, -3)$, so the minimum value is $-3$.
$2$ is the $x$-value where the minimum occurs, not the minimum value itself.
$$f(x)=2(x-3)2+5$$
What is the minimum value of the given function?
Respuesta correcta: D. $5$
The function is in vertex form $a(x-h)2+k$ with $a=2>0$, so the graph opens upward and the vertex $(3, 5)$ is the lowest point. Since $(x-3)2\ge 0$, $f(x)\ge 5$, and $f(3)=5$. The minimum value is $5$.
$3$ is the $x$-coordinate of the vertex, $-3$ has the wrong sign as well, and $2$ is the leading coefficient.
The graph of $y=f(x)$ is shown, where $f(x)=ax2+bx+c$ and $a$, $b$, and $c$ are constants. What is the value of $a+b+c$?
Respuesta numérica (escribes el número).
Respuesta: -8
For any quadratic $f(x)=ax2+bx+c$, $f(1)=a+b+c$. So the question asks for the $y$-value of the graph at $x=1$, and the graph passes through $(1, -8)$. So $a+b+c=-8$.
Check algebraically: the zeros are $-1$ and 5, so $f(x)=a(x+1)(x-5)$. The vertex $(2, -9)$ gives $a(3)(-3)=-9$, so $a=1$ and $f(x)=x2-4x-5$. Then $a+b+c=1-4-5=-8$. Answers such as $-9$ (the vertex’s $y$-value) or $-5$ (the $y$-intercept, which is $c$ alone) read the wrong point.