Linear, Quadratic, and Exponential Functions – SAT Algebra (Calculator)

Solution

In slope-intercept form, \( y = mx + b \), \( m \) is the slope, and \( b \) is the y-intercept.
Here, \( f(x) = 2x + 3 \) means \( m = 2 \) and \( b = 3 \).

Answer: The slope is 2, and the y-intercept is 3.

Solution

First, find the slope: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2. \] Next, use point-slope form with one of the points, say (2,5). The point-slope form is: \[ y - y_1 = m(x - x_1). \] Substituting \( (x_1, y_1) = (2,5) \) and \( m = 2 \): \[ y - 5 = 2(x - 2). \] Expand and simplify to slope-intercept form: \[ y - 5 = 2x - 4 \quad \Rightarrow \quad y = 2x + 1. \]

Answer: \( y = 2x + 1 \).

Solution

Start with \( 3x - 4y = 12 \). Solve for \( y \): \[ -4y = -3x + 12 \quad \Rightarrow \quad y = \frac{-3x + 12}{-4} = \frac{3x}{4} - 3. \] So the slope-intercept form is \( y = \frac{3}{4}x - 3 \).

Hence, the slope is \( \frac{3}{4} \) and the y-intercept is -3.

Answer: \( y = \frac{3}{4}x - 3 \), slope \( = \frac{3}{4} \), intercept \( = -3 \).

Solution

A quick approach is to use the vertex formula for a quadratic \( ax^2 + bx + c \). The x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = 1 \), \( b = -6 \), \( c = 8 \).
So \( x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3. \)
Substitute \( x = 3 \) back into \( f(x) \) to find the y-value: \[ f(3) = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -9 + 8 = -1. \] So the vertex is \( (3, -1) \). Because \( a = 1 > 0 \), the parabola opens upward.

Answer: The vertex is \((3, -1)\), and the parabola opens upward.

Solution

Since \( f(x) = 2^x \), then \( f(3) = 2^3 = 8 \). Because the base 2 is greater than 1, the exponential function \( 2^x \) is increasing for all real x.

Answer: \( f(3) = 8 \), and \( f(x) \) is increasing.

Solution

Use the point-slope form: \( y - y_1 = m(x - x_1) \). Here, \( m = \frac{5}{2} \), \( (x_1, y_1) = (4, 1) \): \[ y - 1 = \frac{5}{2}(x - 4). \] Expand and simplify: \[ y - 1 = \frac{5}{2}x - 10 \] \[ y = \frac{5}{2}x - 10 + 1 \] \[ y = \frac{5}{2}x - 9. \]

Answer: \( y = \frac{5}{2}x - 9 \).

Solution

Because the coefficient of \( x^2 \) is -4, which is negative, the parabola opens downward.
The axis of symmetry for a quadratic \( ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). Here, \( a = -4 \) and \( b = 3 \). \[ x_{\text{axis}} = -\frac{3}{2 \cdot (-4)} = -\frac{3}{-8} = \frac{3}{8}. \]

Answer: The parabola opens downward, and the axis of symmetry is \( x = \frac{3}{8} \).

Solution

An exponential function \( a \cdot b^x \) grows if \( b > 1 \) and decays if \( 0 < b < 1 \). Here, \( b = 0.8 \), which is less than 1, so it represents an exponential decay. The coefficient 5 is the initial value, but since the base is 0.8, the function is decreasing as x increases.

Answer: It models decay because 0.8 < 1.

Solution

The slope \( m = \frac{\Delta y}{\Delta x} = \frac{0 - 8}{4 - 0} = \frac{-8}{4} = -2. \)
One way to write the equation is slope-intercept form. We know the y-intercept is 8 directly from the point (0,8): \[ y = -2x + 8. \]

Answer: The slope is -2, and the equation is \( y = -2x + 8 \).

Solution

Solve \( x^2 - 4x + 3 = 0 \). Factor the left side if possible: \[ x^2 - 4x + 3 = (x - 3)(x - 1). \] Set each factor = 0: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3, \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1. \] Hence, the roots are x = 1 and x = 3, and the factored form is \( (x - 3)(x - 1) \).

Answer: Roots at x=1 and x=3; \( f(x) = (x - 3)(x - 1) \).

Solution

For \( P(t) = 100 \cdot 1.05^t \), the initial value (i.e., at \( t = 0 \)) is \( 100 \cdot 1.05^0 = 100 \cdot 1 = 100 \).
The base of the exponent 1.05 indicates a 5% growth rate per year (since it’s 1 + 0.05).

Answer: Initial value is \$100, and the annual growth rate is 5%.

Solution

In the form \( g(x) = a(x - h)^2 + k \), the vertex is at (h, k). But note the sign in \((x + 1)^2\) is \((x - (-1))^2\). So \( h = -1 \), \( k = 5 \).
Because \( a = -2 \) is negative, the parabola opens downward.

Answer: Vertex is \((-1, 5)\), parabola opens downward.

Solution

A linear model with a constant rate of change of 120 per day can be written as: \[ P(d) = 800 + 120d, \] where 800 is the initial number (day 0), and 120d is the daily increase.

Answer: \( P(d) = 800 + 120d \).

Solution

We can attempt to factor \( 2x^2 - 5x + 2 \). The product of a*c = 2*2 = 4. We want two numbers that multiply to 4 and sum to -5 (the coefficient of x, but negative). Actually, let's systematically factor: \[ 2x^2 - 5x + 2 = (2x - 1)(x - 2)? \] Check: (2x)(x) = 2x^2, outer: (2x)(-2)=-4x, inner: (-1)(x)=-x => total -5x. Constant: (-1)(-2)=2. That works.
Set each factor = 0: \[ 2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2}, \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2. \] These are exact solutions. In decimal form, x=0.5 or x=2.

Answer: \( x = \frac{1}{2} \) and \( x = 2 \).

Solution

Because the exponent’s coefficient (0.2) is positive, \( e^{0.2x} \) increases as x increases. Multiplying by 3 is just a positive scaling, so overall the function is increasing.
When \( x = 0 \), \( y = 3 e^{0.2 \cdot 0} = 3 e^0 = 3(1) = 3.

Answer: It’s an increasing exponential. At \( x=0 \), \( y=3 \).

Solution

The slope of \( y = \frac{2}{3}x - 7 \) is \( \frac{2}{3} \). A line perpendicular to it has slope \( m_{\perp} = -\frac{1}{m} \), specifically the negative reciprocal. The negative reciprocal of \( \frac{2}{3} \) is \( -\frac{3}{2} \).
Now we want a line with slope \( -\frac{3}{2} \) passing through (0,2). Using slope-intercept form, the y-intercept is 2. So \[ y = -\frac{3}{2}x + 2. \]

Answer: Perpendicular slope is \(-\frac{3}{2}\). The line is \( y = -\frac{3}{2}x + 2 \).

Solution

Vertex: We can use the formula \( x = -\frac{b}{2a} \). Here, \( a=4 \), \( b=-16 \), \( c=15 \). \[ x_{\text{vertex}} = -\frac{-16}{2 \cdot 4} = \frac{16}{8} = 2. \] Substituting \( x=2 \) back into the function: \[ f(2) = 4(2^2) - 16(2) + 15 = 4(4) - 32 + 15 = 16 - 32 + 15 = -16 + 15 = -1. \] So the vertex is \((2, -1)\).
y-intercept: This occurs when \( x=0 \). \[ f(0) = 4(0)^2 -16(0) + 15 = 15. \] So the y-intercept is (0,15).

Answer: Vertex at (2, -1), y-intercept at (0, 15).

Solution

Doubling every 5 years means that after 5 years, we multiply by 2, after 10 years, multiply by 2^2, etc. A standard formula is: \[ Q(t) = Q_0 \cdot 2^{\frac{t}{5}}, \] where \( Q_0 = 200 \). So \[ Q(t) = 200 \cdot 2^{\frac{t}{5}}. \]

Answer: \( Q(t) = 200 \cdot 2^{t/5} \).

Solution

Put \( 6x - 2y = 8 \) into slope-intercept form to find its slope. \[ -2y = -6x + 8 \quad \Rightarrow \quad y = 3x - 4. \] The slope is 3. Any line parallel to this line must also have slope 3.

Answer: Slope is 3.

Solution

The standard factor form for a parabola with roots at 1 and 5 is: \[ f(x) = a(x - 1)(x - 5). \] Because it opens upward, \( a > 0 \). The simplest integer coefficient choice is \( a=1 \): \[ f(x) = (x - 1)(x - 5) = x^2 - 6x + 5. \] This is a valid function with the desired roots, opening upward.

Answer: One option is \( f(x) = (x-1)(x-5) = x^2 - 6x + 5. \)

Solution

We want \( A(t) = 600 \cdot e^{kt} \). We know that after 4 years, \( t=4 \), \( A(4)=300 \): \[ 300 = 600 \cdot e^{k \cdot 4}. \] Divide both sides by 600: \[ \frac{300}{600} = e^{4k} \quad \Rightarrow \quad \frac{1}{2} = e^{4k}. \] Take natural log: \[ \ln\left(\frac{1}{2}\right) = 4k \quad \Rightarrow \quad k = \frac{\ln\left(\frac{1}{2}\right)}{4} = \frac{-\ln(2)}{4}. \] So the model is \[ A(t) = 600 \cdot e^{\left(\frac{-\ln(2)}{4}\right)t}. \]

Answer: \( A(t) = 600\,e^{-\frac{\ln(2)}{4}t}. \)

Solution

Let \( m \) = number of markers. We have two points: (4, 5) and (6, 7).
Slope: \[ \frac{7 - 5}{6 - 4} = \frac{2}{2} = 1. \] So the slope is 1. To find the y-intercept (but note that here, the “b” might be the cost with zero markers, which might or might not make sense physically, but we’ll do the math anyway). Let’s use point-slope form with (4, 5): \[ C - 5 = 1(m - 4). \] \[ C - 5 = m - 4 \] \[ C = m + 1. \] This implies that if you buy 0 markers, cost is \$1, which might be a base cost, but mathematically, that’s the linear function that passes through those points.

Answer: \( C(m) = m + 1. \)

Solution

Complete the square: \[ x^2 + 10x + 29 = x^2 + 10x + 25 + 4 = (x + 5)^2 + 4. \] So, in vertex form, that’s \( (x + 5)^2 + 4 \). The vertex is at \((-5, 4)\).
Because the constant term in the vertex form is +4, the minimum value of the parabola is 4, which is above the x-axis. So there are no real zeros (the discriminant will be negative).

Answer: Vertex at \((-5, 4)\). No real zeros, because the parabola never crosses the x-axis.

Solution

Point-slope form: \[ y - 7 = -3(x - 2). \] Expand: \[ y - 7 = -3x + 6. \] So slope-intercept form: \[ y = -3x + 6 + 7 = -3x + 13. \]

Answer: \( y - 7 = -3(x - 2) \) or \( y = -3x + 13 \).

Solution

y-intercept occurs at \( x=0 \). Evaluate: \[ p(0) = -(0)^2 + 2(0) + 8 = 8. \] So the point is (0,8).
Maximum point: For a parabola \( ax^2 + bx + c \) with \( a<0 \), the vertex is a maximum. Use \( x= -\frac{b}{2a} \). Here, \( a=-1 \), \( b=2 \). \[ x_{\text{vertex}} = -\frac{2}{2 \cdot (-1)} = -\frac{2}{-2} = 1. \] Evaluate p(1): \[ p(1) = -(1)^2 + 2(1) + 8 = -1 + 2 + 8 = 9. \] So the vertex (maximum) is (1, 9).

Answer: y-int at (0,8). Maximum at (1,9).

Solution

We want \( N(t) = 200 \). So \[ 100 \cdot (1.1)^t = 200. \] Divide both sides by 100: \[ (1.1)^t = 2. \] Take log (base 10 or natural) and solve for t: \[ t \ln(1.1) = \ln(2) \quad \Rightarrow \quad t = \frac{\ln(2)}{\ln(1.1)}. \] Numerically, \(\ln(2) \approx 0.6931\), \(\ln(1.1) \approx 0.09531\). So \( t \approx \frac{0.6931}{0.09531} \approx 7.27 \) years.

Answer: About 7.27 years until the population reaches 200.

Solution

We have points (2,10) and (6,18).
Slope: \[ m = \frac{18-10}{6-2} = \frac{8}{4} = 2. \] Using slope-intercept form or point-slope form: \[ y - 10 = 2(x - 2) \quad \Rightarrow \quad y - 10 = 2x - 4 \quad \Rightarrow \quad y=2x+6. \]

Answer: \( L(x) = 2x + 6. \)

Solution

Recall that \( a^x = e^{x \ln(a)} \). For \( y = 3^x \), we can rewrite it as: \[ y = e^{x \ln(3)}. \]

Answer: \( y = e^{( \ln(3) )x}. \)

Solution

If the roots are -2 and 5, in factored form we have: \[ q(x) = a(x + 2)(x - 5). \] Next, use \( q(0)=10 \) to find a. Evaluate: \[ q(0) = a(0 + 2)(0 - 5) = a \cdot 2 \cdot (-5) = -10a. \] We want this to be 10, so \[ -10a = 10 \quad \Rightarrow \quad a = -1. \] Thus, \[ q(x) = -1(x+2)(x-5). \] Expand: \[ q(x) = -(x^2 - 5x + 2x - 10) = -(x^2 - 3x - 10) = -x^2 + 3x + 10. \]

Answer: \( q(x) = -x^2 + 3x + 10. \)

Solution

Substitute \( x=-2 \): \[ f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15. \] 15 is positive, so the point \((-2,15)\) is above the x-axis.

Answer: \( f(-2)=15 \), which is above the x-axis.

Solution

The factor \(\left(\frac{1}{2}\right)^{\frac{t}{6}}\) means the substance halves every 6 years.
Specifically, after 6 years: \[ R(6) = 100 \left(\frac{1}{2}\right)^{\frac{6}{6}} = 100 \left(\frac{1}{2}\right)^1 = 50. \]

Answer: Every 6 years, the substance is half of its previous amount; at t=6, R(6)=50.

Solution

Solve for y: \[ 2y = 4x + 12 \quad \Rightarrow \quad y=2x + 6. \] So slope = 2, y-intercept = 6.

Answer: Slope=2, intercept=6.

Solution

We have \( y=a(x-2)^2+1 \). Plug in (x,y)=(3,4): \[ 4 = a(3-2)^2 +1 = a(1)^2 +1 = a +1. \] So \( a=3. \)
Thus, the function is \( y=3(x-2)^2+1 \).

Answer: \( a=3 \), so \( y=3(x-2)^2+1 \).

Solution

In logarithmic form, \( x = \log_{5}(20). \) We can use natural log or common log: \[ x = \frac{\ln(20)}{\ln(5)}. \] Numerically, \(\ln(20) \approx 2.9957\), \(\ln(5)\approx1.6094\). \[ x \approx \frac{2.9957}{1.6094} \approx 1.86. \]

Answer: \( x=\log_5(20)\approx1.86.\)

Solution

Expand: \[ (x+1)(x-4)=x(x-4)+1(x-4)=x^2-4x+x-4=x^2-3x-4. \] So the standard form is \( x^2 -3x -4 \).
The x-coordinate of the vertex is \(-\frac{b}{2a} = -\frac{-3}{2\cdot1}=\frac{3}{2}.\)

Answer: \( x^2-3x-4 \). Vertex at \( x=\frac{3}{2}.\)

Solution

For \( ax^2 + bx + c=0 \), \( x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \)
Here, a=3, b=2, c=-1. \[ x=\frac{-2\pm\sqrt{(2)^2-4(3)(-1)}}{2\cdot3} = \frac{-2\pm\sqrt{4+12}}{6} = \frac{-2\pm\sqrt{16}}{6} = \frac{-2\pm4}{6}. \] So we get \( x=\frac{-2+4}{6}=\frac{2}{6}=\frac{1}{3} \) or \( x=\frac{-2-4}{6}=\frac{-6}{6}=-1. \)

Answer: \( x=\frac{1}{3} \) or \( x=-1.\)

Solution

Shifting a function \( y=f(x) \) up by 5 means we add 5 to the output: \[ y_{\text{new}} = 4x-2+5 = 4x+3. \]

Answer: \( y_{\text{new}}=4x+3.\)

Solution

Recall \( 2^{x+1}=2^x\cdot2^1=2\cdot 2^x. \) So \( h(x)=2\cdot2^x.\)

Answer: \( h(x)=2\cdot2^x.\)

Solution

y-intercept: set x=0 => \( y=-2(0)^2+12(0)-5=-5.\) So the y-intercept is (0,-5).
Vertex x-value: a=-2, b=12 => x= -b/(2a)= -12/(2*(-2))= -12/(-4)=3.

Answer: y-int= (0,-5). Vertex x=3.

Solution

Rewrite: \( 4^{x}=25 \). Taking natural log: \[ x \ln(4)=\ln(25) \quad => \quad x=\frac{\ln(25)}{\ln(4)}. \] Numerically, \(\ln(25)\approx3.2189,\ln(4)\approx1.3863,\) so \( x\approx 3.2189/1.3863\approx2.32.\)

Answer: \( x=\frac{\ln(25)}{\ln(4)} \approx2.32.\)

Solution

Factor: \( x^2+6x+5=(x+1)(x+5). \) Set each factor=0 => x=-1, x=-5 => The x-intercepts are (-1,0) and (-5,0).

Answer: \( f(x)=(x+1)(x+5)\); x-intercepts x=-1, -5.

Solution

The slope of the line \( y=-2x+5\) is -2. A parallel line has the same slope. Use point-slope with (3,2): \[ y-2=-2(x-3) => y-2=-2x+6 => y=-2x+8. \]

Answer: \( y=-2x+8.\)

Solution

\( B(0)=50\cdot e^{0.3\cdot0}=50\cdot e^0=50.\)
For t=10, \[ B(10)=50\cdot e^{0.3\times10}=50\cdot e^{3}. \] Numerically, \( e^3\approx20.0855,\) => \( B(10)\approx 50\times20.0855=1004.275.\)

Answer: \( B(0)=50,\; B(10)\approx1004.\)

Solution

Expand \((x-2)^2=x^2-4x+4.\) \[ p(x)=3(x^2-4x+4)+4=3x^2-12x+12+4=3x^2-12x+16. \] The vertex is from the form \((x-2)^2\), so vertex= (2,4).

Answer: \( p(x)=3x^2-12x+16,\) with vertex (2,4).

Solution

For exponential growth, we need \( b>1.\) For example, \( g(x)=100\cdot1.05^x\) grows by 5% each increment of x, because 1.05>1.

Answer: \( b>1\) ensures growth. E.g., \( 1.05^x\) grows.

Solution

\( x^2-9=0 => x^2=9.\) The solutions are \( x=\pm3.\)
Alternatively, we can factor: \( x^2-9=(x-3)(x+3)=0.\)

Answer: \( x=\pm3.\)

Solution

Slope= \(\frac{7-4}{2-1}=\frac{3}{1}=3.\) Because the slope is positive, the line is increasing from left to right.

Answer: Slope=3, line is increasing.

Solution

\(\text{x-intercepts}\): from factors => x=1 or x=-2 => points (1,0),(-2,0).
\(\text{y-intercept}\): set x=0 => y=(0-1)(0+2)= -1\cdot2= -2 => (0,-2).

Answer: x-ints at (1,0) and (-2,0), y-int at (0,-2).

Solution

Logarithmic form: \( x=\log_7(50).\) Using change of base, \[ x=\frac{\ln(50)}{\ln(7)}\approx\frac{3.9120}{1.9459}\approx2.01. \]

Answer: \( x=\log_7(50)\approx2.01.\)

Solution

The function is in vertex form \( -3(x-4)^2+2.\)
Vertex: (4,2).
Axis of symmetry: \( x=4.\)
Because the coefficient -3 is negative, the parabola opens downward, so it has a maximum at the vertex.
For the y-intercept, set x=0: \[ y=-3(0-4)^2+2=-3(16)+2=-48+2=-46. \]

Answer: Vertex at (4,2). Axis: x=4. It has a maximum. y-int is (0,-46).