EBUS PC11 Chapter 3 Quadratic Functions Test-1

Step 1: The vertex of a parabola is the highest or lowest point on the graph. Step 2: From the graph, the lowest point occurs at \(x=-3\) and \(y=-2\). Step 3: Therefore, the vertex is \((-3,-2)\). Answer: C. \((-3, -2)\)

Step 1: A quadratic function has a parabola graph that continues forever to the left and right. Step 2: Therefore, the domain is all real numbers. Step 3: The parabola opens downward, and its highest point is \(y=9\). Step 4: So the range is \(y \le 9\). Answer: A. D: all real numbers; R: \(y \le 9\)

Step 1: The vertex form is \(y=a(x-p)^2+q\). Step 2: For \(y=\frac{1}{2}(x-3)^2+5\), the value of \(p\) is \(3\). Step 3: The axis of symmetry is the vertical line \(x=p\). Step 4: Therefore, the axis of symmetry is \(x=3\). Answer: D. \(x=3\)

Step 1: Rewrite the function as \(f(x)=2x^2+12x+16\). Step 2: To find the x-intercepts, set \(f(x)=0\): \(2x^2+12x+16=0\). Step 3: Factor out the GCF: \(2(x^2+6x+8)=0\). Step 4: Factor the trinomial: \(2(x+2)(x+4)=0\). Step 5: Set each factor equal to zero: \(x+2=0\) or \(x+4=0\). Step 6: So \(x=-2\) or \(x=-4\). Step 7: To find the y-intercept, substitute \(x=0\): \(f(0)=16\). Answer: D. \(x: -2, -4;\ y: 16\)

Step 1: A vertical expansion by a factor of \(3\) means \(a=3\). Step 2: A translation 2 units right means \(p=2\) in \(y=a(x-p)^2+q\). Step 3: A translation 4 units down means \(q=-4\). Step 4: Therefore, \(a=3\), \(p=2\), and \(q=-4\). Answer: C. \(a=3,\ p=2,\ q=-4\)

Step 1: The equation is in vertex form \(y=a(x-p)^2+q\). Step 2: For \(y=-\frac{1}{2}(x-2)^2+4\), the vertex is \((2,4)\). Step 3: Since \(a=-\frac{1}{2}\), the parabola opens downward. Step 4: The correct graph must have vertex \((2,4)\) and open downward. Step 5: Graph C matches these features. Answer: C. Graph C

Step 1: The vertex form is \(y=a(x-h)^2+k\). Step 2: The vertex is \(V(3,-3)\), so \(h=3\) and \(k=-3\). Step 3: The parabola opens up and is congruent to \(y=\frac{1}{3}x^2\), so \(a=\frac{1}{3}\). Step 4: Substitute these values: \(y=\frac{1}{3}(x-3)^2-3\). Answer: B. \(y=\frac{1}{3}(x-3)^2-3\)

Step 1: Use vertex form \(y=a(x-h)^2+k\). Step 2: The vertex is \(V(-5,0)\), so \(h=-5\) and \(k=0\). Step 3: The equation becomes \(y=a(x+5)^2\). Step 4: The y-intercept is \(-50\), so substitute \(x=0\) and \(y=-50\): \(-50=a(0+5)^2\). Step 5: This gives \(-50=25a\), so \(a=-2\). Step 6: Therefore, the equation is \(y=-2(x+5)^2\). Answer: C. \(y=-2(x+5)^2\)

Step 1: Use vertex form \(y=a(x-h)^2+k\). Step 2: The vertex is \((1,3)\), so the equation is \(y=a(x-1)^2+3\). Step 3: The parabola passes through \((3,5)\), so substitute \(x=3\) and \(y=5\): \(5=a(3-1)^2+3\). Step 4: Simplify: \(5=4a+3\). Step 5: Subtract 3: \(2=4a\). Step 6: Divide by 4: \(a=\frac{1}{2}\). Step 7: Therefore, the equation is \(y=\frac{1}{2}(x-1)^2+3\). Answer: D. \(y=\frac{1}{2}(x-1)^2+3\)

Step 1: Start with \(y=x^2-7x-20\). Step 2: Take half of \(-7\), which is \(-\frac{7}{2}\), and square it: \(\left(-\frac{7}{2}\right)^2=\frac{49}{4}\). Step 3: Add and subtract \(\frac{49}{4}\): \(y=x^2-7x+\frac{49}{4}-\frac{49}{4}-20\). Step 4: Rewrite the perfect square: \(y=\left(x-\frac{7}{2}\right)^2-\frac{49}{4}-20\). Step 5: Convert \(-20\) to fourths: \(-20=-\frac{80}{4}\). Step 6: Combine constants: \(-\frac{49}{4}-\frac{80}{4}=-\frac{129}{4}\). Step 7: So \(y=\left(x-\frac{7}{2}\right)^2-\frac{129}{4}\). Answer: D. \(y=\left(x-\frac{7}{2}\right)^2-\frac{129}{4}\)

Step 1: Start with \(y=-x^2-4x-3\). Step 2: Factor out \(-1\) from the quadratic and linear terms: \(y=-(x^2+4x)-3\). Step 3: Complete the square inside the brackets: \(x^2+4x=(x+2)^2-4\). Step 4: Substitute: \(y=-\left((x+2)^2-4\right)-3\). Step 5: Distribute the negative sign: \(y=-(x+2)^2+4-3\). Step 6: Simplify: \(y=-(x+2)^2+1\). Answer: D. \(y=-(x+2)^2+1\)

Step 1: To find x-intercepts, set \(f(x)=0\): \(x^2-8x-4=0\). Step 2: Complete the square: \(x^2-8x=(x-4)^2-16\). Step 3: Substitute: \((x-4)^2-16-4=0\). Step 4: Simplify: \((x-4)^2-20=0\). Step 5: Add 20 to both sides: \((x-4)^2=20\). Step 6: Take the square root of both sides: \(x-4=\pm\sqrt{20}\). Step 7: Simplify: \(\sqrt{20}=2\sqrt{5}\). Step 8: Therefore, \(x=4\pm2\sqrt{5}\). Answer: A. \(4\pm2\sqrt{5}\)

Step 1: Start with \(y=x^2-8x-4\). Step 2: Complete the square: \(x^2-8x=(x-4)^2-16\). Step 3: Substitute into the function: \(y=(x-4)^2-16-4\). Step 4: Simplify: \(y=(x-4)^2-20\). Step 5: Since the coefficient of \((x-4)^2\) is positive, the parabola opens upward. Step 6: The vertex is \((4,-20)\), so the minimum value is \(-20\). Answer: A. Minimum value of \(-20\)

Step 1: The profit function is a quadratic \(p=-10x^2+300x-250\). Step 2: Since \(a=-10<0\), the parabola opens downward, so the vertex gives the maximum profit. Step 3: Use the vertex formula \(x=\frac{-b}{2a}\). Step 4: Here, \(a=-10\) and \(b=300\). Step 5: Calculate \(x=\frac{-300}{2(-10)}=\frac{-300}{-20}=15\). Step 6: Therefore, the selling price that gives the maximum profit is \(\$15\). Answer: A. \(\$15\)

Step 1: Let \(x\) be the width of each of the two fenced sides perpendicular to the river. Step 2: Then the remaining fenced side has length \(100-2x\). Step 3: The area is \(A=x(100-2x)=100x-2x^2\). Step 4: This is a quadratic opening downward, so the maximum occurs at the vertex. Step 5: Use \(x=\frac{-b}{2a}\) with \(a=-2\) and \(b=100\). Step 6: \(x=\frac{-100}{2(-2)}=25\). Step 7: The length is \(100-2(25)=50\). Step 8: The maximum area is \(25\cdot50=1250\text{ m}^2\). Answer: C. \(1250\text{ m}^2\)

Step 1: Let \(n\) be the number of \(\$5\) price increases. Step 2: The price is \(40+5n\). Step 3: The number of subscriptions sold is \(2000-200n\). Step 4: The revenue is \(R=(40+5n)(2000-200n)\). Step 5: Expand: \(R=80000+2000n-1000n^2\). Step 6: This quadratic opens downward, so the maximum occurs at the vertex. Step 7: Use \(n=\frac{-b}{2a}\), where \(a=-1000\) and \(b=2000\). Step 8: \(n=\frac{-2000}{2(-1000)}=1\). Step 9: The best price is \(40+5(1)=45\). Answer: C. \(\$45\)