Question

when does the graph of the function ( h(x) = x^4 + 3x^2 - 4 ) cross the x - axis?
a. ( x = - 2, x = 2 )
b. ( x = - 2, x = - 1, x = 2 )
c. ( x = - 2, x = - 1, x = 1 )
d. ( x = - 1, x = 1 )

Explanation:

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

We need to find when the graph of \( h(x)=x^{4}+3x^{2}-4 \) crosses the \( x \)-axis, so we set \( h(x) = 0 \), which gives the equation \( x^{4}+3x^{2}-4 = 0 \).

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

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

Step3: Solve the quadratic equation

For a quadratic equation \( ay^{2}+by + c = 0 \) (here \( a = 1 \), \( b = 3 \), \( c=-4 \)), we can use the quadratic formula \( y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) or factor it. Factoring: \( y^{2}+3y - 4=(y + 4)(y - 1)=0 \). So, \( y+4 = 0 \) or \( y - 1=0 \), which gives \( y=-4 \) or \( y = 1 \).

Step4: Substitute back \( y = x^{2} \)

• For \( y=-4 \): \( x^{2}=-4 \), which has no real solutions since the square of a real number cannot be negative.
• For \( y = 1 \): \( x^{2}=1 \), then \( x=\pm1 \) (since if \( x^{2}=1 \), then \( x = \sqrt{1}=1 \) or \( x=-\sqrt{1}=-1 \)). Wait, but let's check the original options. Wait, maybe I misread the function. Wait, the function is \( h(x)=x^{4}+3x^{2}-4 \)? Wait, no, maybe it's \( h(x)=x^{4}+3x^{3}-4 \)? No, the user's image shows \( h(x)=x^{4}+3x^{2}-4 \)? Wait, the options have \( x = - 2,x = 2,x=-1,x = 1 \). Wait, maybe the function is \( h(x)=x^{4}+3x^{3}-4x^{2} \)? No, let's re - evaluate. Wait, the options: Let's check the options again. The options are:

A. \( x=-2,x = 2 \)

B. \( x=-2,x=-1,x = 2 \)

C. \( x=-2,x=-1,x = 1 \)

D. \( x=-1,x = 1 \)

Wait, maybe the function is \( h(x)=x^{4}+3x^{3}-4x^{2} \)? No, let's factor the given function \( h(x)=x^{4}+3x^{2}-4 \) correctly. Wait, \( x^{4}+3x^{2}-4=(x^{2}+4)(x^{2}-1)=(x^{2}+4)(x - 1)(x + 1) \). So the real roots are \( x = 1 \) and \( x=-1 \), because \( x^{2}+4>0 \) for all real \( x \). But the options have C as \( x=-2,x=-1,x = 1 \) and D as \( x=-1,x = 1 \). Wait, maybe the function is \( h(x)=x^{4}+3x^{3}-4x^{2} \). Let's try that. If \( h(x)=x^{4}+3x^{3}-4x^{2}=x^{2}(x^{2}+3x - 4)=x^{2}(x + 4)(x - 1) \). Then roots are \( x = 0 \) (double root), \( x=-4 \), \( x = 1 \). Not matching. Wait, maybe the function is \( h(x)=x^{4}+3x^{2}-4x \)? No. Wait, maybe the original function is \( h(x)=x^{4}+3x^{3}-4x \). Then \( h(x)=x(x^{3}+3x^{2}-4)=x(x^{3}-x^{2}+4x^{2}-4)=x[x^{2}(x - 1)+4(x^{2}-1)]=x[x^{2}(x - 1)+4(x - 1)(x + 1)]=x(x - 1)(x^{2}+4x + 4)=x(x - 1)(x + 2)^{2} \). Then roots are \( x = 0,x = 1,x=-2 \). But the options have C: \( x=-2,x=-1,x = 1 \), D: \( x=-1,x = 1 \). Wait, maybe the function is \( h(x)=x^{4}+3x^{2}-4 \) is wrong, and the correct function is \( h(x)=x^{4}+3x^{3}-4x^{2}-12x \). Let's factor: \( h(x)=x(x^{3}+3x^{2}-4x - 12)=x[x^{2}(x + 3)-4(x + 3)]=x(x + 3)(x^{2}-4)=x(x + 3)(x - 2)(x + 2) \). Then roots are \( x = 0,x=-3,x = 2,x=-2 \). Not matching.

Wait, maybe the user made a typo, but based on the function \( h(x)=x^{4}+3x^{2}-4=(x^{2}+4)(x - 1)(x + 1) \), the real roots are \( x = 1 \) and \( x=-1 \), which is option D. But let's check the options again. Option D is \( x=-1 \) and \( x = 1 \).

Answer:

D. \( x=-1 \) and \( x = 1 \)