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Quadratic Functions

Sol 1 The x-coordinate is given by $x=-\frac{b}{2a}=-6/(2*1)=-3$, and then the y-coordinate is given by $y=(-3)^2+6(-3)-2=9-18-2=-11$.

Sol 2 The x-coordinate of the vertex is given by $x=-\frac{b}{2a}=-(-5)/(2(-1))=-5/2$; so (since $a<0$) the maximum value is given by $y=-(-5/2)^2-5(-5/2)+9=-25/4+25/2+9=25/4+9=61/4$.

Sol 3 Since 5 and 1 are the x-intercepts of the graph, $f(x)=a(x-5)(x-1)$ for some number $a$. Since the graph of $f$ is symmetric around the vertical line passing through its vertex, the x-coordinate of the vertex is 3 (the average of 5 and 1). Therefore the minimum value of $f$ is $f(3)=a(3-5)(3-1)=-4a=-12$, so $a=3$ and $f(x)=3(x-5)(x-1)$ or $f(x)=3(x^2-6x+5)=3x^2-18x+15$.

Sol 4 Since the vertex is at $(-2,3)$, the parabola has an equation of the form $y-3=a(x-(-2))^2$ or $y-3=a(x+2)^2$. Since the parabola passes through the point $(2,-1)$, substituting $x=2$ and $y=-1$ gives $-4=a(4^2)=16a$, so $a=-1/4$ and $y-3=-1/4(x+2)^2=-1/4(x^2+4x+4)=-1/4x^2-x-1$, so $y=(-1/4)x^2-x+2$.

Sol 5 Substituting $t=x^2$ in $g(x)$ and replacing $g(x)$ by $f(t)$ gives the quadratic function $f(t)=t^2-8t+25$. Then the t-coordinate of the vertex of the graph of $f$ is given by $t=-\frac{b}{2a}=-(-8)/(2*1)=4$, so the minimum value of $f$ is given by $f(4)=4^2-8(4)+25=16-32+25=9$ Therefore the minimum value of $g$ is also 9, and this value is attained when $x^2=t=4$ so $x=\pm 2$.

Sol 6 Since the roles of $x$ and $y$ are interchanged, the y-coordinate of the vertex is given by $y=-\frac{b}{2a}=-(-10)/(2*1)=5$ and the x-coordinate is given by $x=5^2-10^5+40=25-50+40=15$.

Sol 7 Since $f(2)=-20$ is the minimum value of $f$, the graph of $f$ has an equation of the form $y-(-20)=a(x-2)^2$ or $y+20=a(x-2)^2$ with $a>0$. Since $f(4)=-8$, substituting $x=4$ and $y=-8$ gives $12=a(2^2)=4a$ and therefore $a=3$. Therefore $y+20=3(x-2)^2$, so $f(x)=3(x-2)^2-20$ or $f(x)=3(x^2-4x+4)-20=3x^2-12x-8$.

Sol 8 Since the line is non-vertical and passes through the point $(3,9)$, it has an equation of the form $y-9=m(x-3)$ or $y=mx-3m+9$. Since the line intersects the parabola $y=x^2$ exactly once, the equation $x^2=mx-3m+9$ or $x^2-mx+(3m-9)=0$ has exactly one solution. Therefore the discriminant $b^2-4ac$ is equal to 0, so $(-m)^2-4(1)(3m-9)=0\Rightarrow m^2-12m+36=0$ $\Rightarrow (m-6)^2=0\Rightarrow m-6=0 \Rightarrow m=6$,so the line has equation $y-9=6(x-3)$ or $y=6x-9$.

Sol 9 (If the horizontal line $y=k$ intersects the graph of $f$ only once, then $k$ will correspond to the maximum value or minimum value of $f$.) Setting $f(x)=k$ gives $\frac{12x}{x^2+4}=k$, so $12x=kx^2+4k$ and therefore $0=kx^2-12x+4k$. This equation will have exactly one solution if the discriminant is zero, so $b^2-4ac=(-12)^2-4(k)(4k)=0\Rightarrow 144-16k^2=0$ $\Rightarrow 144=16k^2\Rightarrow k^2=9 \Rightarrow k=\pm3$. Therefore $f$ has a maximum value of 3, since $k>3\Rightarrow k^2>9\Rightarrow b^2-4ac=144-16k^2<0 \Rightarrow f(x)=k$ has no solution.



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Lawrence Marx 2002-07-13