The function is defined by What is the minimum value of ?

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SAT Nonlinear Functions Question 218 | Free SAT Prep | Aniko

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Question 218·Medium·Nonlinear Functions

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The function $h$ is defined by

h(t) = (t - 4)^2 + (t - 4) + 10

What is the minimum value of $h(t)$ ?

When asked for the minimum or maximum of a quadratic on the SAT, first recognize it as a parabola of the form $ax^2 + bx + c$ . If the expression is shifted, consider a substitution (like $x = t - 4$ ) to simplify. Then quickly convert to vertex form by completing the square, or use the vertex formula $x = -\dfrac{b}{2a}$ and plug this x-value back into the function. Because the squared term is always nonnegative, the constant in the vertex form directly gives the minimum (if $a>0$ ) or maximum (if $a<0$ ), letting you choose the correct answer efficiently without graphing.

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Recognize the type of function

Look at the highest power of $t$ in $h(t) = (t - 4)^2 + (t - 4) + 10$ . What kind of function is this, and what does that tell you about whether it has a maximum or a minimum?

Simplify using a substitution

Try letting $x = t - 4$ to rewrite $h(t)$ in terms of $x$ . This can make it easier to see the structure of the quadratic.

Use vertex/complete-the-square idea

For a quadratic like $x^2 + x + 10$ , you can find the minimum by completing the square or by using the vertex formula $x = -\dfrac{b}{2a}$ and then plugging back in. Focus on rewriting it as a perfect square plus a constant.

Desmos Guide

Enter the function

In Desmos, type h(t) = (t - 4)^2 + (t - 4) + 10 so that the graph of the function appears.

View the graph and locate the lowest point

Look at the parabola on the graph. Since it opens upward, its minimum value occurs at its vertex (the lowest point on the curve). Zoom in if needed so the vertex is clearly visible.

Read the minimum value

Click or tap on the vertex of the parabola, or use the minimum(h, a, b) function over an interval that clearly includes the vertex (for example, minimum(h, 0, 10)). The y-coordinate of that minimum point is the minimum value of $h(t)$ .

Step-by-step Explanation

Rewrite the function in simpler form

Notice that $h(t)$ depends on $t$ only through $(t-4)$ . Let $x = t - 4$ to simplify the expression.

Then

h(t) = (t - 4)^2 + (t - 4) + 10 = x^2 + x + 10.

So the problem becomes: What is the minimum value of $x^2 + x + 10$ ?

Complete the square for the quadratic

To find the minimum of $x^2 + x + 10$ , complete the square for the $x^2 + x$ part.

Take half of the coefficient of $x$ (which is $1$ ), square it, and add and subtract that value:

x^2 + x + 10 = x^2 + x + \left(\frac12\right)^2 + 10 - \left(\frac12\right)^2 \\[0.7em] = (x + \tfrac12)^2 + 10 - \tfrac14.

Now combine the constants $10$ and $-\tfrac14$ .

Identify the minimum value from the completed square form

Combine the constants:

10 - \frac14 = \frac{40}{4} - \frac14 = \frac{39}{4}.

So we can write the function as

h(t) = x^2 + x + 10 = (x + \tfrac12)^2 + \frac{39}{4},

where $x = t - 4$ .

The term $(x + \tfrac12)^2$ is a square, so it is always greater than or equal to $0$ , and its smallest possible value is $0$ . Therefore, the smallest possible value of $h(t)$ occurs when $(x + \tfrac12)^2 = 0$ , and the minimum value of $h(t)$ is $\dfrac{39}{4}$ , which corresponds to choice B.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation