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Interpret slope and intercept in context

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5290379
A market analyst models the supply price for a product by \(p(x)=1.5x+20\), where \(x\) is the number of units and \(p(x)\) is the price per unit in dollars. A \(\$5\) subsidy per unit is paid directly to producers, so producers can request \(\$5\) less from buyers for the same total payment. a) Write the new supply-price function \(p_{\mathrm{sub}}(x)\). b) Explain why the subsidy is represented by a downward vertical shift. c) Find and interpret the new y-intercept within this model.

Hints

- Determine how much buyers must pay after the producer receives the subsidy. - Subtracting the same constant from every output creates a vertical shift. - Evaluate the new function at \(x=0\). - Interpret the intercept only within the stated model.

Solution

1. Subtract the subsidy from every modeled price: \(p_{\mathrm{sub}}(x)=p(x)-5=1.5x+15\). 2. Because every output decreases by the same amount, every point on the graph moves down \(5\) units. 3. At \(x=0\), \(p_{\mathrm{sub}}(0)=15\), so the y-intercept is \((0, 15)\). 4. Within the algebraic model, the intercept means the modeled price is \(\$15\) when the quantity is \(0\). It should not automatically be interpreted as a real-world minimum operating price without additional economic assumptions.

Answer

a) \(p_{\mathrm{sub}}(x)=1.5x+15\) b) The subsidy subtracts \(5\) from every output, so the graph shifts down \(5\) units. c) The y-intercept is \((0, 15)\), representing a modeled price of \(\$15\) at a quantity of \(0\).
5288179
In a geothermal model, underground temperature increases approximately linearly with depth. The temperature is \(25\,\text{°C}\) at a depth of \(500\,\text{m}\) and \(70\,\text{°C}\) at a depth of \(2000\,\text{m}\). a) Write a linear function \(T(d)\) for temperature as a function of depth \(d\), measured in meters. b) What surface temperature does the model predict? c) At what depth does the model predict a temperature of \(100\,\text{°C}\)?

Hints

- Treat depth as the input and temperature as the output. - Use the two data points to find the rate of change. - The y-intercept represents the model's temperature at depth \(0\). - To find a depth from a temperature, set the function equal to that temperature and solve for \(d\).

Solution

1. Use the points \((500, 25)\) and \((2000, 70)\). The slope is \(m=\frac{70-25}{2000-500}=\frac{45}{1500}=0.03\) degrees Celsius per meter. 2. Substitute \((500, 25)\) into \(T(d)=0.03d+b\): \(25=0.03\cdot 500+b\), so \(b=10\). Therefore, \(T(d)=0.03d+10\). 3. At the surface, \(d=0\), so \(T(0)=10\). The predicted surface temperature is \(10\,\text{°C}\). 4. Set \(T(d)=100\): \(100=0.03d+10\). Thus, \(90=0.03d\), so \(d=3000\,\text{m}\).

Answer

a) \(T(d)=0.03d+10\) b) \(10\,\text{°C}\) c) \(3000\,\text{m}\)
5288279
An energy company offers two annual residential plans. The Flex plan has a base charge of \(\$120\) and an energy rate of \(6.0\) cents per kWh. The Active plan has a base charge of \(\$180\) and an energy rate of \(4.5\) cents per kWh. 1. Find the annual usage at which the two plans cost the same. 2. The company raises the Active plan's base charge by \(20\%\). Find the new break-even usage and state when Active becomes less expensive. 3. Explain how lowering the Active plan's energy rate would affect the intersection of the two cost graphs, assuming all other values remain unchanged.

Hints

- Convert cents per kWh to dollars per kWh before writing the functions. - Equal costs correspond to equal function values. - A percentage increase in a base charge changes the y-intercept. - A lower energy rate produces a smaller slope.

Solution

1. In dollars, the cost functions are \(F(x)=0.06x+120\) and \(A(x)=0.045x+180\). Set them equal: \(0.06x+120=0.045x+180\). Then \(0.015x=60\), so \(x=4000\). The plans cost the same at \(4000\,\text{kWh}\). 2. The new Active base charge is \(180\cdot 1.20=\$216\). Solve \(0.06x+120=0.045x+216\): \(0.015x=96\), so \(x=6400\). Active is less expensive for \(x>6400\,\text{kWh}\). 3. Lowering Active's energy rate decreases the slope of its cost graph. Because its base charge remains higher, the graphs intersect at a lower usage value, so Active becomes advantageous sooner.

Answer

1. \(4000\,\text{kWh}\) 2. New break-even usage: \(6400\,\text{kWh}\); Active costs less for \(x>6400\,\text{kWh}\). 3. The intersection moves left, to a lower usage value.
5288389
Two monthly phone plans are modeled by linear functions, where \(x\) is data usage in gigabytes and \(K(x)\) is the monthly cost in dollars. Plan A: \(K_A(x)=2x+20\) Plan B: \(K_B(x)=3x+5\) 1. Identify \(m\) and \(b\) for each plan and explain what each parameter means in context. 2. Determine when Plan A costs less than Plan B.

Hints

- In \(K(x)=mx+b\), identify the part that changes with usage and the fixed part. - Use the units to interpret the slope and intercept. - Write a strict inequality to represent “Plan A costs less.”

Solution

1. For Plan A, \(m=2\) and \(b=20\). The rate is \(\$2\) per gigabyte, and the monthly base charge is \(\$20\). 2. For Plan B, \(m=3\) and \(b=5\). The rate is \(\$3\) per gigabyte, and the monthly base charge is \(\$5\). 3. Compare the plans: \(2x+20<3x+5\). Subtracting \(2x\) and \(5\) gives \(15<x\). Therefore, Plan A costs less when usage exceeds \(15\) GB.

Answer

1. Plan A: \(m=2\) dollars per GB, \(b=\$20\). Plan B: \(m=3\) dollars per GB, \(b=\$5\). 2. Plan A costs less when \(x>15\) GB.
5288289
An energy company models a family of annual plans by \(K_p(x)=px+(300-2000p)\), where \(x\) is annual usage in kWh, \(p\) is the price per kWh in dollars, and \(K_p(x)\) is the annual cost in dollars. 1. Show that all graphs in the family pass through one common point. Find the point and interpret it in context. 2. A customer uses \(3500\,\text{kWh}\) per year. Determine whether a higher or lower value of \(p\) gives this customer a lower annual cost.

Hints

- Factor the parameter \(p\) from the function expression. - Find an x-value that makes the coefficient of \(p\) equal to \(0\). - For part 2, substitute \(3500\) and analyze the resulting expression as a function of \(p\).

Solution

1. Rewrite the family as \(K_p(x)=p(x-2000)+300\). At \(x=2000\), the parameter term is zero, so \(K_p(2000)=300\) for every value of \(p\). The common point is \((2000, 300)\). At annual usage of \(2000\,\text{kWh}\), every plan in the family costs \(\$300\). 2. Substitute \(x=3500\): \(K_p(3500)=3500p+300-2000p=1500p+300\). The coefficient of \(p\) is positive, so the cost increases as \(p\) increases. A lower value of \(p\) is less expensive for this customer.

Answer

1. Common point: \((2000, 300)\). At \(2000\,\text{kWh}\), every plan costs \(\$300\). 2. A lower price per kWh is less expensive at \(3500\,\text{kWh}\).

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