SAT MAth Practice questions – all topics
- Advanced Math Weightage: 35% Questions: 13-15
- Equivalent expressions
- Nonlinear equations in one variable and systems of equations in two variables
- Nonlinear functions
SAT MAth and English – full syllabus practice tests
Question Hard
\[
f(x)=(2 x+3)(2 x-5)
\]
What is the minimum value of the given function?
A) -16
B) \(-\frac{3}{2}\)
C) \(\frac{1}{2}\)
D) 2
▶️Answer/Explanation
Ans: A
To find the minimum value of the function \( f(x) = (2x + 3)(2x – 5) \), we first expand it into a standard quadratic form.
\[
f(x) = (2x + 3)(2x – 5)
\]
\[
f(x) = 4x^2 – 10x + 6x – 15
\]
\[
f(x) = 4x^2 – 4x – 15
\]
The quadratic function \( f(x) = 4x^2 – 4x – 15 \) is in the standard form \( ax^2 + bx + c \), where \( a = 4 \), \( b = -4 \), and \( c = -15 \).
The minimum value of a quadratic function \( ax^2 + bx + c \) occurs at the vertex. The \(x\)-coordinate of the vertex can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = 4 \) and \( b = -4 \):
\[
x = -\frac{-4}{2 \times 4} = \frac{4}{8} = \frac{1}{2}
\]
Next, substitute \( x = \frac{1}{2} \) back into the function to find the corresponding \( y \)-value:
\[
f\left( \frac{1}{2} \right) = 4 \left( \frac{1}{2} \right)^2 – 4 \left( \frac{1}{2} \right) – 15
\]
\[
f\left( \frac{1}{2} \right) = 4 \cdot \frac{1}{4} – 2 – 15
\]
\[
f\left( \frac{1}{2} \right) = 1 – 2 – 15
\]
\[
f\left( \frac{1}{2} \right) = -16
\]
Therefore, the minimum value of the function is:
\[
\boxed{-16}
\]
Question Hard
For the quadratic function \(h\), the table gives three values of \(x\) and their corresponding values of \(h(x)\). At what value of \(x\) does \(h\) reach its minimum?
A. -1
B. 0
C. 3
D. 4
▶️Answer/Explanation
Ans:C
To determine at what value of \(x\) the quadratic function \(h(x)\) reaches its minimum, we need to look for the vertex of the parabola defined by the table.
Given the values of \(h(x)\) for \(x = 2\), \(x = 4\), and \(x = 6\), we can notice that the function \(h(x)\) has a symmetric shape, typical of a quadratic function. Since the values at \(x = 2\) and \(x = 4\) are both \(0\), and the value at \(x = 6\) is \(8\), we can infer that the vertex of the parabola lies between \(x = 2\) and \(x = 6\).
The vertex of a quadratic function in the form \(h(x) = ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\). However, since we don’t have the exact equation of \(h(x)\), we can still find the \(x\)-coordinate of the vertex by taking the average of the \(x\)-values where \(h(x)\) is \(0\). In this case, that would be the average of \(2\) and \(4\), which is \(3\).
Therefore, the minimum value of \(h(x)\) occurs at \(x = 3\). So, the correct answer is option C: \(3\).
Question Hard
$
x^2-4 x-9=0
$
The solutions to the given equation can be written in the form \(\frac{\mathrm{m} \pm \sqrt{\mathrm{k}}}{2}\), where \(\mathrm{m}\) and \(\mathrm{k}\) are intergers. What is the value of \(\mathrm{m}+\mathrm{k}\) ?
▶️Answer/Explanation
Ans: 56
To find the solutions to the quadratic equation \(x^2 – 4x – 9 = 0\), we can use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]
Where \(a = 1\), \(b = -4\), and \(c = -9\).
Substitute the values into the formula:
\[x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(-9)}}{2(1)}\]
\[x = \frac{4 \pm \sqrt{16 + 36}}{2}\]
\[x = \frac{4 \pm \sqrt{52}}{2}\]
The solutions can be written as \(m\sqrt{k}\), where \(m = 4\) and \(k = 52\). Therefore, \(m + k = 4 + 52 = 56\).
Question Hard
\[x^2+2 x-1=0\]
A solution to the given equation is \(\sqrt{k}-1\). What is the value of \(k\) ?
▶️Answer/Explanation
2
Given the equation \(x^2 + 2x – 1 = 0\) and a solution \(\sqrt{k} – 1\), let’s substitute this solution into the equation:
\[ (\sqrt{k} – 1)^2 + 2(\sqrt{k} – 1) – 1 = 0 \]
Expanding and simplifying:
\[ k – 2\sqrt{k} + 1 + 2\sqrt{k} – 2 – 1 = 0 \]
\[ k – 2 = 0 \]
So, \(k = \boxed{2}\).
Question Hard
When the quadratic function \(f\) is graphed in
the \(x y\)-plane, where \(y=f(x)\), its vertex is \((-2,5)\). One of the \(x\)-intercepts of this graph is \(\left(-\frac{7}{3}, 0\right)\). What is the other \(x\)-intercept of the graph?
A. \(\left(-\frac{13}{3}, 0\right)\)
B. \(\left(-\frac{5}{3}, 0\right)\)
C. \(\left(\frac{1}{3}, 0\right)\)
D. \(\left(\frac{7}{3}, 0\right)\)
▶️Answer/Explanation
Ans:B