Chapter 3
Algebra and Trigonometry with Analytic Geometry · 370 exercises
Problem 25
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{x+1}{x^{3}-4 x} $$
6 step solution
Problem 25
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ (x+3)^{2}+(y-2)^{2}=9 $$
3 step solution
Problem 25
Find a formula that expresses the fact that \(P(x, y)\) is a distance 5 from the origin. Describe the set of all such points.
5 step solution
Problem 26
Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=-\sqrt{16-(c x)^{2}} ; \quad c=1, \frac{1}{2}, 4 $$
6 step solution
Problem 26
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\sqrt{3-x}, \quad g(x)=\sqrt{x+2} $$
5 step solution
Problem 26
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(0,-2) ; \quad \text { slope } 5 $$
5 step solution
Problem 26
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{4 x}{6 x^{2}+13 x-5} $$
4 step solution
Problem 26
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ (x-4)^{2}+(y+2)^{2}=4 $$
4 step solution
Problem 26
Find a formula that states that \(P(x, y)\) is a distance \(r>0\) from a fixed point \(C(h, k)\). Describe the set of all such points.
6 step solution
Problem 27
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\sqrt{3-x}, \quad g(x)=\sqrt{x^{2}-16} $$
7 step solution
Problem 27
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(4,-5) ; \quad \text { through } B(-3,6) $$
5 step solution
Problem 27
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{\sqrt{2 x-3}}{x^{2}-5 x+4} $$
3 step solution
Problem 27
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ (x+3)^{2}+y^{2}=16 $$
5 step solution
Problem 27
Find a formula that states that \(P(x, y)\) is a distance \(r>0\) from a fixed point \(C(h, k)\). Describe the set of all such points.
5 step solution
Problem 28
Exer. 27-32: If the point \(P\) is on the graph of a function \(f\), find the corresponding point on the graph of the given function. $$ P(3,-1) ; \quad y=2 f(x)+4 $$
4 step solution
Problem 28
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=x^{3}+5, \quad g(x)=\sqrt[3]{x-5} $$
6 step solution
Problem 28
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(-1,6) ; \quad x \text {-intercept } 5 $$
5 step solution
Problem 28
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ x^{2}+(y-2)^{2}=25 $$
4 step solution
Problem 28
Find all points on the \(x\)-axis that are a distance 5 from \(P(-2,4)\).
5 step solution
Problem 29
Exer. 27-32: If the point \(P\) is on the graph of a function \(f\), find the corresponding point on the graph of the given function. $$ P(3,-2) ; \quad y=2 f(x-4)+1 $$
4 step solution
Problem 29
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{3 x+5}{2}, \quad g(x)=\frac{2 x-5}{3} $$
5 step solution
Problem 29
Exer. 29-34: Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. Vertex \((0,-2)\), passing through \((3,25)\)
6 step solution
Problem 29
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(2,-4) ; \text { parallel to the line } 5 x-2 y=4 $$
6 step solution
Problem 29
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{\sqrt{4 x-3}}{x^{2}-4} $$
3 step solution
Problem 29
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ 4 x^{2}+4 y^{2}=25 $$
4 step solution
Problem 29
Find the point with coordinates of the form \((2 a, a)\) that is in the third quadrant and is a distance 5 from \(P(1,3)\).
7 step solution
Problem 30
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{1}{x-1}, \quad g(x)=x-1 $$
4 step solution
Problem 30
Exer. 29-34: Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. Vertex \((0,5)\), passing through \((2,-3)\)
5 step solution
Problem 30
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(-3,5) ; \text { parallel to the line } x+3 y=1 $$
4 step solution
Problem 30
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{1}{(x-3) \sqrt{x+3}} $$
4 step solution
Problem 30
Find all points with coordinates of the form \((a, a)\) that are a distance 3 from \(P(-2,1)\).
7 step solution
Problem 31
Exer. 27-32: If the point \(P\) is on the graph of a function \(f\), find the corresponding point on the graph of the given function. $$ P(3,9) ; \quad y=\frac{1}{3} f\left(\frac{1}{2} x\right)-1 $$
6 step solution
Problem 31
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=x^{2}, \quad g(x)=\frac{1}{x^{3}} $$
5 step solution
Problem 31
Exer. 29-34: Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. Vertex \((3,5), x\)-intercept 0
4 step solution
Problem 31
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(7,-3) ; \quad \text { perpendicular to the line } 2 x-5 y=8 $$
4 step solution
Problem 31
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\sqrt{x+2}+\sqrt{2-x} $$
5 step solution
Problem 31
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ y=-\sqrt{16-x^{2}} $$
6 step solution
Problem 31
For what values of \(a\) is the distance between \(P(a, 3)\) and \(Q(5,2 a)\) greater than \(\sqrt{26}\) ?
7 step solution
Problem 32
Exer. 27-32: If the point \(P\) is on the graph of a function \(f\), find the corresponding point on the graph of the given function. $$ P(-2,1) ; \quad y=-3 f(2 x)-5 $$
5 step solution
Problem 32
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{x}{x-2}, \quad g(x)=\frac{3}{x} $$
4 step solution
Problem 32
Exer. 29-34: Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. Vertex \((4,-7), x\)-intercept \(-4\)
5 step solution
Problem 32
Exer. 21-32: Find a general form of an equation of the line through the point \(A\) that satisfies the given condition. $$ A(4,5) ; \quad \text { perpendicular to the line } 3 x+2 y=7 $$
6 step solution
Problem 32
Exer. 21-32: Find the domain of \(f\). $$ f(x)=\sqrt{(x-2)(x-6)} $$
5 step solution
Problem 32
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ y=\sqrt{4-x^{2}} $$
5 step solution
Problem 32
Given \(A(-2,0)\) and \(B(2,0)\), find a formula not containing radicals that expresses the fact that the sum of the distances from \(P(x, y)\) to \(A\) and to \(B\), respectively, is 5 .
7 step solution
Problem 33
Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{x-1}{x-2}, \quad g(x)=\frac{x-3}{x-4} $$
4 step solution
Problem 33
Exer. 29-34: Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. \(x\)-intercepts \(-3\) and 5 , highest point has \(y\)-coordinate 4
4 step solution
Problem 33
Exer. 33-36: Find the slope-intercept form of the line that satisfies the given conditions. $$ x \text {-intercept } 4, \quad y \text {-intercept }-3 $$
4 step solution
Problem 33
Exer. 23-34: Sketch the graph of the circle or semicircle. $$ x=\sqrt{9-y^{2}} $$
4 step solution
Problem 33
Prove that the midpoint of the hypotenuse of any right triangle is equidistant from the vertices.
7 step solution