SAT Parabola Problems: Min/Max and Vertex Applications
Practice & Review
Apply your knowledge of parabola forms and features to solve SAT-style word problems. Focus on interpreting questions about maximum height, minimum cost, and other real-world scenarios modeled by quadratic functions.
SAT Parabola Problems: Min/Max and Vertex Applications
1 of 12
In an SAT word problem, what keywords suggest that you need to find the vertex of a parabola?
Look for words like "maximum," "minimum," "highest," "lowest," "greatest," "least," "peak," and "bottom." These usually refer to the y-coordinate of the vertex.
A question asks for the "time at which a projectile reaches its maximum height." Are you looking for the x-coordinate or the y-coordinate of the vertex?
You are looking for the x-coordinate of the vertex, as "time" is typically represented on the x-axis.
A question asks for the "maximum height of a projectile." Are you looking for the x-coordinate or the y-coordinate of the vertex?
You are looking for the y-coordinate of the vertex, as "height" is represented on the y-axis and this is the maximum value of the function.
The height \(h\), in feet, of a ball thrown upwards is given by the equation \(h(t) = -16t^2 + 64t + 4\), where \(t\) is the time in seconds. What is the maximum height the ball reaches?
This asks for the y-coordinate of the vertex. First find the time (x-coord): \(t = -\frac{64}{2(-16)} = 2\) seconds. Then find the height: \(h(2) = -16(2)^2 + 64(2) + 4 = -64 + 128 + 4 = 68\) feet.
For the equation \(h(t) = -16t^2 + 64t + 4\), how long does it take for the ball to reach its highest point?
This asks for the x-coordinate (in this case, t-coordinate) of the vertex. \(t = -\frac{b}{2a} = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2\) seconds.
The profit, \(P\), from selling an item is given by \(P(x) = -0.1x^2 + 50x - 1000\), where \(x\) is the number of items sold. How many items must be sold to maximize profit?
This asks for the x-coordinate of the vertex. \(x = -\frac{50}{2(-0.1)} = -\frac{50}{-0.2} = 250\). To maximize profit, \(250\) items must be sold.
Using the profit function \(P(x) = -0.1x^2 + 50x - 1000\), what is the maximum possible profit?
This asks for the y-coordinate of the vertex. We know \(x = 250\) maximizes profit. \(P(250) = -0.1(250)^2 + 50(250) - 1000 = -6250 + 12500 - 1000 = 5250\). The maximum profit is \($5250\).
A parabola has x-intercepts at \(x = -2\) and \(x = 8\). What is the x-coordinate of its vertex?
The vertex is always halfway between the x-intercepts. The x-coordinate of the vertex is the average of the intercepts: \(\frac{-2 + 8}{2} = \frac{6}{2} = 3\). So, the x-coordinate is \(3\).
The equation \(y = (x-20)(x+4)\) represents a parabola. At what x-value does the minimum value of \(y\) occur?
The x-intercepts are at \(x=20\) and \(x=-4\). The x-coordinate of the vertex (where the minimum occurs) is the average of the intercepts: \(\frac{20 + (-4)}{2} = \frac{16}{2} = 8\). The minimum occurs at \(x=8\).
For the parabola \(y = a(x-r_1)(x-r_2)\), what is the equation for its axis of symmetry?
The axis of symmetry is the vertical line passing through the midpoint of the roots \(r_1\) and \(r_2\). The equation is \(x = \frac{r_1 + r_2}{2}\).
A parabola has a vertex at \((3, 5)\) and passes through the point \((1, -3)\). The equation is of the form \(y = a(x-h)^2 + k\). Find the value of \(a\).
We know \(h=3\) and \(k=5\), so \(y = a(x-3)^2 + 5\). Plug in the point \((1, -3)\) for \(x\) and \(y\): \(-3 = a(1-3)^2 + 5\). This simplifies to \(-3 = a(-2)^2 + 5\), so \(-8 = 4a\), which means \(a = -2\).
The height of an arch is modeled by \(y = -0.5x^2 + 4x\), where \(x\) is the horizontal distance from the base. What is the width of the arch at its base?
The width is the distance between the x-intercepts. Find the intercepts by setting \(y=0\): \(0 = -0.5x^2 + 4x = x(-0.5x + 4)\). The intercepts are \(x=0\) and \(-0.5x + 4 = 0 \Rightarrow x = 8\). The width is \(8 - 0 = 8\) units.
