Solved on Nov 09, 2023

Find the possible values of $k$ for the linear function $g(x)=x+k$ that make the quadratic function $f(x)=2x^2+1$ intersect with $g(x)$ twice, once, or not at all.

STEP 1

Assumptions1. The quadratic function is $f(x)=x^{}+1$ . The linear function is $g(x)=x+k$ , where $k$ is a real number3. We are looking for the values of $k$ such that $f(x)$ and $g(x)$ intersect twice, once, or not at all

STEP 2

The point of intersection between two functions $f(x)$ and $g(x)$ is the value of $x$ for which $f(x) = g(x)$ . So, we set $f(x)$ equal to $g(x)$ and solve for $x$ .
$2x^{2}+1 = x+k$

STEP 3

Rearrange the equation to form a quadratic equation.
$2x^{2} - x + (1 - k) =0$

STEP 4

The solutions of a quadratic equation $ax^{2} + bx + c =0$ are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

STEP 5

In our equation, $a =2$ , $b = -1$ , and $c =1 - k$ . The discriminant of the quadratic equation, which determines the number of solutions, is $b^{2} -4ac$ .

STEP 6

Calculate the discriminant.
$= (-1)^{2} -4 \cdot2 \cdot (1 - k)$

STEP 7

implify the discriminant.
$=1 - +k$

STEP 8

Further simplify the discriminant.
$=8k -7$

STEP 9

The number of intersections between $f(x)$ and $g(x)$ depends on the discriminant $$.
a. If $>$ , the quadratic equation has two distinct real roots, so $f(x)$ and $g(x)$ intersect twice. b. If $=$ , the quadratic equation has exactly one real root, so $f(x)$ and $g(x)$ intersect once. c. If $<$ , the quadratic equation has no real roots, so $f(x)$ and $g(x)$ do not intersect.

STEP 10

For $f(x)$ and $g(x)$ to intersect twice, $>0$ . So we solve the inequality $8k -7 >0$ .

STEP 11

Rearrange the inequality to solve for $k$ .
$8k >7$

STEP 12

Divide by8 to isolate $k$ .
$k > \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to intersect twice, $k$ must be greater than $\frac{7}{8}$ .

STEP 13

For $f(x)$ and $g(x)$ to intersect once, $=0$ . So we solve the equation $8k -7 =0$ .

STEP 14

Rearrange the equation to solve for $k$ .
$8k =7$

STEP 15

Divide by8 to isolate $k$ .
$k = \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to intersect once, $k$ must be equal to $\frac{7}{8}$ .

STEP 16

For $f(x)$ and $g(x)$ to not intersect, $<0$ . So we solve the inequality $8k - <0$ .

STEP 17

Rearrange the inequality to solve for $k$ .
$k <7$

STEP 18

Divide by8 to isolate $k$ .
$k < \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to not intersect, $k$ must be less than $\frac{7}{8}$ .

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Find the possible values of $k$ for the linear function $g(x)=x+k$ that make the quadratic function $f(x)=2x^2+1$ intersect with $g(x)$ twice, once, or not at all.

STEP 1

Assumptions1. The quadratic function is $f(x)=x^{}+1$ . The linear function is $g(x)=x+k$ , where $k$ is a real number3. We are looking for the values of $k$ such that $f(x)$ and $g(x)$ intersect twice, once, or not at all

STEP 2

The point of intersection between two functions $f(x)$ and $g(x)$ is the value of $x$ for which $f(x) = g(x)$ . So, we set $f(x)$ equal to $g(x)$ and solve for $x$ .
$2x^{2}+1 = x+k$

STEP 3

Rearrange the equation to form a quadratic equation.
$2x^{2} - x + (1 - k) =0$

STEP 4

The solutions of a quadratic equation $ax^{2} + bx + c =0$ are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

STEP 5

In our equation, $a =2$ , $b = -1$ , and $c =1 - k$ . The discriminant of the quadratic equation, which determines the number of solutions, is $b^{2} -4ac$ .

STEP 6

Calculate the discriminant.
$= (-1)^{2} -4 \cdot2 \cdot (1 - k)$

STEP 7

implify the discriminant.
$=1 - +k$

STEP 8

Further simplify the discriminant.
$=8k -7$

STEP 9

STEP 10

For $f(x)$ and $g(x)$ to intersect twice, $>0$ . So we solve the inequality $8k -7 >0$ .

STEP 11

Rearrange the inequality to solve for $k$ .
$8k >7$

STEP 12

Divide by8 to isolate $k$ .
$k > \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to intersect twice, $k$ must be greater than $\frac{7}{8}$ .

STEP 13

For $f(x)$ and $g(x)$ to intersect once, $=0$ . So we solve the equation $8k -7 =0$ .

STEP 14

Rearrange the equation to solve for $k$ .
$8k =7$

STEP 15

Divide by8 to isolate $k$ .
$k = \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to intersect once, $k$ must be equal to $\frac{7}{8}$ .

STEP 16

For $f(x)$ and $g(x)$ to not intersect, $<0$ . So we solve the inequality $8k - <0$ .

STEP 17

Rearrange the inequality to solve for $k$ .
$k <7$

STEP 18

Divide by8 to isolate $k$ .
$k < \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to not intersect, $k$ must be less than $\frac{7}{8}$ .

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Find the possible values of kkk for the linear function g(x)=x+kg(x)=x+kg(x)=x+k that make the quadratic function f(x)=2x2+1f(x)=2x^2+1f(x)=2x2+1 intersect with g(x)g(x)g(x) twice, once, or not at all.

STEP 1

Assumptions1. The quadratic function is f(x)=x+1f(x)=x^{}+1f(x)=x+1 . The linear function is g(x)=x+kg(x)=x+kg(x)=x+k, where kkk is a real number3. We are looking for the values of kkk such that f(x)f(x)f(x) and g(x)g(x)g(x) intersect twice, once, or not at all

STEP 2

The point of intersection between two functions f(x)f(x)f(x) and g(x)g(x)g(x) is the value of xxx for which f(x)=g(x)f(x) = g(x)f(x)=g(x). So, we set f(x)f(x)f(x) equal to g(x)g(x)g(x) and solve for xxx.2x2+1=x+k2x^{2}+1 = x+k2x2+1=x+k

STEP 3

Rearrange the equation to form a quadratic equation.2x2−x+(1−k)=02x^{2} - x + (1 - k) =02x2−x+(1−k)=0

STEP 4

The solutions of a quadratic equation ax2+bx+c=0ax^{2} + bx + c =0ax2+bx+c=0 are given by the quadratic formulax=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}x=2a−b±b2−4ac​​

STEP 5

In our equation, a=2a =2a=2, b=−1b = -1b=−1, and c=1−kc =1 - kc=1−k. The discriminant of the quadratic equation, which determines the number of solutions, is b2−4acb^{2} -4acb2−4ac.

STEP 6

Calculate the discriminant.=(−1)2−4⋅2⋅(1−k) = (-1)^{2} -4 \cdot2 \cdot (1 - k)=(−1)2−4⋅2⋅(1−k)

STEP 7

implify the discriminant.=1−+k =1 - +k=1−+k

STEP 8

Further simplify the discriminant.=8k−7 =8k -7=8k−7

STEP 9

STEP 10

For f(x)f(x)f(x) and g(x)g(x)g(x) to intersect twice, >0 >0>0. So we solve the inequality 8k−7>08k -7 >08k−7>0.

STEP 11

Rearrange the inequality to solve for kkk.8k>78k >78k>7

STEP 12

Divide by8 to isolate kkk.k>78k > \frac{7}{8}k>87​So, for f(x)f(x)f(x) and g(x)g(x)g(x) to intersect twice, kkk must be greater than 78\frac{7}{8}87​.

STEP 13

For f(x)f(x)f(x) and g(x)g(x)g(x) to intersect once, =0 =0=0. So we solve the equation 8k−7=08k -7 =08k−7=0.

STEP 14

Rearrange the equation to solve for kkk.8k=78k =78k=7

STEP 15

Divide by8 to isolate kkk.k=78k = \frac{7}{8}k=87​So, for f(x)f(x)f(x) and g(x)g(x)g(x) to intersect once, kkk must be equal to 78\frac{7}{8}87​.

STEP 16

For f(x)f(x)f(x) and g(x)g(x)g(x) to not intersect, <0 <0<0. So we solve the inequality 8k−<08k - <08k−<0.

STEP 17

Rearrange the inequality to solve for kkk.k<7k <7k<7

STEP 18

Divide by8 to isolate kkk.k<78k < \frac{7}{8}k<87​So, for f(x)f(x)f(x) and g(x)g(x)g(x) to not intersect, kkk must be less than 78\frac{7}{8}87​.

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Find the possible values of $k$ for the linear function $g(x)=x+k$ that make the quadratic function $f(x)=2x^2+1$ intersect with $g(x)$ twice, once, or not at all.

Assumptions1. The quadratic function is $f(x)=x^{}+1$ . The linear function is $g(x)=x+k$ , where $k$ is a real number3. We are looking for the values of $k$ such that $f(x)$ and $g(x)$ intersect twice, once, or not at all

The point of intersection between two functions $f(x)$ and $g(x)$ is the value of $x$ for which $f(x) = g(x)$ . So, we set $f(x)$ equal to $g(x)$ and solve for $x$ .
$2x^{2}+1 = x+k$

Rearrange the equation to form a quadratic equation.
$2x^{2} - x + (1 - k) =0$

The solutions of a quadratic equation $ax^{2} + bx + c =0$ are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

In our equation, $a =2$ , $b = -1$ , and $c =1 - k$ . The discriminant of the quadratic equation, which determines the number of solutions, is $b^{2} -4ac$ .

Calculate the discriminant.
$= (-1)^{2} -4 \cdot2 \cdot (1 - k)$

implify the discriminant.
$=1 - +k$

Further simplify the discriminant.
$=8k -7$

For $f(x)$ and $g(x)$ to intersect twice, $>0$ . So we solve the inequality $8k -7 >0$ .

Rearrange the inequality to solve for $k$ .
$8k >7$

Divide by8 to isolate $k$ .
$k > \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to intersect twice, $k$ must be greater than $\frac{7}{8}$ .

For $f(x)$ and $g(x)$ to intersect once, $=0$ . So we solve the equation $8k -7 =0$ .

Rearrange the equation to solve for $k$ .
$8k =7$

Divide by8 to isolate $k$ .
$k = \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to intersect once, $k$ must be equal to $\frac{7}{8}$ .

For $f(x)$ and $g(x)$ to not intersect, $<0$ . So we solve the inequality $8k - <0$ .

Rearrange the inequality to solve for $k$ .
$k <7$

Divide by8 to isolate $k$ .
$k < \frac{7}{8}$ So, for $f(x)$ and $g(x)$ to not intersect, $k$ must be less than $\frac{7}{8}$ .