Question

question 1-1
where does the graph of the function ( h(x) = x^4 + 3x^2 - 4 ) cross the x-axis?
( circ ) ( x = -2, x = 2 )
( circ ) ( x = -2, x = -1, and x = 2 )
( circ ) ( x = -2, x = -1, and x = 1 )
( circ ) ( x = -1 and x = 1 )

Explanation:

Step1: Set \( h(x) = 0 \)

To find where the graph crosses the \( x \)-axis, we set \( h(x)=0 \). So we have the equation \( x^{4}+3x^{2}-4 = 0 \).

Step2: Substitute \( u = x^{2} \)

Let \( u=x^{2} \), then the equation becomes a quadratic equation in terms of \( u \): \( u^{2}+3u - 4=0 \).

Step3: Solve the quadratic equation for \( u \)

We factor the quadratic equation \( u^{2}+3u - 4 = 0 \). We need two numbers that multiply to \( - 4 \) and add up to \( 3 \). The numbers are \( 4 \) and \( - 1 \). So, \( u^{2}+3u - 4=(u + 4)(u - 1)=0 \).
Setting each factor equal to zero:
- For \( u + 4=0 \), we get \( u=-4 \). But since \( u = x^{2}\geq0 \) for all real \( x \), \( u=-4 \) gives no real solutions.
- For \( u - 1=0 \), we get \( u = 1 \).

Step4: Solve for \( x \) when \( u = 1 \)

Since \( u=x^{2}=1 \), we solve for \( x \) by taking the square root of both sides. \( x=\pm\sqrt{1}=\pm1 \)? Wait, no, wait. Wait, we made a mistake above. Wait, let's re - factor the quadratic. Wait, \( u^{2}+3u - 4=(u + 4)(u - 1) \) is correct. But when \( u = x^{2} \), if we have \( x^{4}+3x^{2}-4 = 0 \), another way:
Wait, maybe we can factor the original polynomial directly. \( x^{4}+3x^{2}-4=(x^{2}+4)(x^{2}-1) \). Then, set \( (x^{2}+4)(x^{2}-1)=0 \).
- \( x^{2}+4 = 0 \) gives \( x^{2}=-4 \), no real solutions.
- \( x^{2}-1=0 \) gives \( x^{2}=1 \), so \( x=\pm1 \)? Wait, but that's not matching the options. Wait, maybe I made a mistake in the substitution. Wait, let's check the original function again. The function is \( h(x)=x^{4}+3x^{2}-4 \). Wait, maybe I factored wrong. Wait, \( x^{4}+3x^{2}-4=(x^{2}+4)(x^{2}-1) \) is correct. But the options have \( x = \pm2 \) and \( x=\pm1 \). Wait, maybe I misread the function. Wait, maybe the function is \( x^{4}-3x^{2}-4 \)? No, the user provided \( h(x)=x^{4}+3x^{2}-4 \). Wait, no, let's re - do the substitution. Wait, if we set \( u = x^{2} \), the equation is \( u^{2}+3u - 4=0 \). The quadratic formula: \( u=\frac{-3\pm\sqrt{9 + 16}}{2}=\frac{-3\pm5}{2} \). So \( u=\frac{-3 + 5}{2}=1 \) or \( u=\frac{-3-5}{2}=-4 \). So \( u = 1 \) gives \( x^{2}=1\Rightarrow x=\pm1 \), \( u=-4 \) gives no real solutions. But the options have \( x=\pm2 \) and \( x=\pm1 \). Wait, maybe the function is \( x^{4}-3x^{2}-4 \)? Let's check. If \( h(x)=x^{4}-3x^{2}-4 \), then set \( x^{4}-3x^{2}-4 = 0 \), let \( u=x^{2} \), \( u^{2}-3u - 4=0 \), factor: \( (u - 4)(u+1)=0 \), \( u = 4 \) or \( u=-1 \). \( u = 4\Rightarrow x^{2}=4\Rightarrow x=\pm2 \), \( u=-1 \) no solution. But the original function is \( x^{4}+3x^{2}-4 \). Wait, there must be a mistake. Wait, looking at the options, the correct approach is:
Wait, maybe the function is \( h(x)=x^{4}-3x^{2}-4 \). Let's assume that (maybe a typo in the problem statement). Then:
Set \( h(x)=0 \), \( x^{4}-3x^{2}-4 = 0 \), let \( u = x^{2} \), \( u^{2}-3u - 4=0 \), factor: \( (u - 4)(u + 1)=0 \), \( u = 4 \) or \( u=-1 \). \( u = 4\Rightarrow x^{2}=4\Rightarrow x=\pm2 \), \( u=-1 \) no solution. But the options have \( x=-2,x = - 1,x = 1,x = 2 \). Wait, maybe the function is \( x^{4}-x^{2}-4 \)? No, let's go back. Wait, the options include \( x=-2,x=-1,x = 1,x = 2 \). So let's test the values.
Test \( x = 2 \): \( h(2)=2^{4}+3\times2^{2}-4=16 + 12-4=24 eq0 \).
Test \( x=-2 \): \( h(-2)=(-2)^{4}+3\times(-2)^{2}-4=16 + 12-4=24 eq0 \).
Test \( x = 1 \): \( h(1)=1+3 - 4=0 \).
Test \( x=-1 \): \( h(-1)=1 + 3-4=0 \).
Wait, but the options have \( x=-2,x=-1,x = 1,x = 2 \) as an option. Wait, maybe the function is \( x^{4}-3x^{2}-4 \). Let's test \( x = 2 \): \( 16-12 - 4=0 \). \( x=-2 \): \( 16-12 - 4=0 \). \( x = 1 \): \( 1-3 - 4=-6 eq0 \). No, that's not. Wait, maybe the original function is \( x^{4}+3x^{2}-4 \), and there is a mistake in the options, or I made a mistake. Wait, the third option is \( x=-2,x=-1,x = 2 \)? No, the options are:
1. \( x=-2,x = 2 \)
2. \( x=-2,x=-1,x = 2 \)
3. \( x=-2,x=-1,x = 1 \)
4. \( x=-1,x = 1 \)

Wait, let's re - evaluate \( h(x)=x^{4}+3x^{2}-4 \) at \( x = 1 \): \( 1 + 3-4=0 \). At \( x=-1 \): \( 1+3 - 4=0 \). At \( x = 2 \): \( 16+12 - 4=24 eq0 \). At \( x=-2 \): \( 16 + 12-4=24 eq0 \). So the correct solutions are \( x=\pm1 \), which is the fourth option. But maybe the function was supposed to be \( x^{4}-3x^{2}-4 \). Let's check \( x = 2 \): \( 16-12 - 4=0 \), \( x=-2 \): \( 16-12 - 4=0 \), \( x = 1 \): \( 1-3 - 4=-6 eq0 \), \( x=-1 \): \( 1-3 - 4=-6 eq0 \). No. Wait, maybe the function is \( x^{4}+3x^{3}-4x \)? No, the user wrote \( h(x)=x^{4}+3x^{2}-4 \). So according to the calculation, when \( h(x) = 0 \), \( x^{4}+3x^{2}-4=(x^{2}+4)(x^{2}-1)=0 \), so \( x^{2}-1=0\Rightarrow x=\pm1 \). So the answer should be \( x=-1 \) and \( x = 1 \), which is the fourth option. But maybe there is a typo in the problem. Assuming the function is correct as given, the solution is \( x=\pm1 \).

Answer:

\( x=-1 \) and \( x = 1 \) (corresponding to the option "x = - 1 and x = 1")